One algebra problem

Chapter 9, “Quadratics” from Beginning and Intermediate Algebra by Tyler Wallace is available under a Creative Commons Attribution 3.0 Unported license . © 2010. 9.1Quadratics - Solving with Radicals Ob jective: Solve equations with radicals and check for extr aneous solu- tions.

Here we look at equations that have roots in the problem. As yo u might expect, to clear a root we can raise both sides to an exponent. So to cle ar a square root we can rise both sides to the second power. To clear a cubed roo t we can raise both sides to a third power. There is one catch to solving a pro blem with roots in it, sometimes we end up with solutions that do not actually wo rk in the equation.

This will only happen if the index on the root is even, and it wi ll not happen all the time. So for these problems it will be required that we che ck our answer in the original problem. If a value does not work it is called an e xtraneous solution and not included in the nal solution.

When solving a radical problem with an even index: check answ ers!

Example 442. 7x + 2 √ = 4 Even index !We will have to check answers ( 7 x+ 2 √ ) 2 = 4 2 Square both sides ,simplify exponents 7 x + 2 = 16 Solve − 2 −2 Subtract 2from both sides 7 x = 14 Divide both sides by 7 7 7 x = 2 Need to check answer in original problem 7(2) + 2 p = 4 Multiply 14 + 2 √ = 4 Add 16 √ = 4 Square root 4 = 4 True!It works !

x = 2 Our Solution Example 443. x− 1 3 √ = − 4 Odd index ,we don ′ t need to check answer ( x− 1 3 √ ) 3 = ( −4)3 Cube both sides ,simplify exponents x − 1 = −64 Solve 326 + 1 + 1 Add1to both sides x = − 63 Our Solution Example 444. 3x + 6 4 √ = − 3 Even index !We will have to check answers ( 3 x+ 6 4 √ ) = ( −3)4 Rise both sides to fourth power 3 x + 6 = 81 Solve − 6 −6 Subtract 6from both sides 3 x = 75 Divide both sides by 3 3 3 x = 25 Need to check answer in original problem 3( 25 ) + 6 4 p = − 3 Multiply 75 + 6 4 √ = − 3 Add 81 4 √ = − 3 Take root 3 = −3 False ,extraneous solution No Solution Our Solution If the radical is not alone on one side of the equation we will h ave to solve for the radical before we raise it to an exponent Example 445.

x+ 4 x+ 1 √ = 5 Even index !We will have to check solutions − x −x Isolate radical by subtracting xfrom both sides 4 x + 1 √ = 5 −x Square both sides ( 4 x+ 1 √ ) 2 = (5 −x)2 Evaluate exponents ,recal (a − b)2 = a2 − 2a b +b2 4 x + 1 = 25−10 x+ x2 Re−order terms 4 x + 1 = x2 − 10 x+ 25 Make equation equal zero − 4x − 1 −4x −1 Subtract 4x and 1from both sides 0 = x2 − 14 x+ 24 Factor 0 = ( x− 12 )(x− 2) Set each factor equal to zero x − 12 = 0 orx− 2 = 0 Solve each equation + 12 +12 + 2 + 2 x= 12 or x= 2 Need to check answers in original problem ( 12 ) + 4( 12) + 1 p = 5 Check x= 5 rst 327 12+ 48+ 1 √ = 5 Add 12 + 49 √ = 5 Take root 12 + 7 = 5 Add 19 = 5 False,extraneous root (2) + 4(2) + 1 p = 5 Check x= 2 2 + 8 + 1 √ = 5 Add 2 + 9 √ = 5 Take root 2 + 3 = 5 Add 5 = 5 True!It works x = 2 Our Solution The above example illustrates that as we solve we could end up with anx2 term or a quadratic. In this case we remember to set the equation to zero and solve by factoring. We will have to check both solutions if the index i n the problem was even. Sometimes both values work, sometimes only one, and so metimes neither works.

World View Note: The babylonians were the rst known culture to solve quadratics in radicals - as early as 2000 BC!

If there is more than one square root in a problem we will clear the roots one at a time. This means we must rst isolate one of them before we squ are both sides.

Example 446. 3x − 8 √ − x √ = 0 Even index !We will have to check answers + x √ + x √ Isolate rst root by adding x √ to both sides 3 x − 8 √ = x √ Square both sides ( 3 x− 8 √ ) 2 = ( x √ ) 2 Evaluate exponents 3 x − 8 = x Solve − 3x −3x Subtract 3x from both sides − 8 = −2x Divide both sides by −2 − 2 −2 4 = x Need to check answer in original 3(4) −8 p − 4 √ = 0 Multiply 12 −8 √ − 4 √ = 0 Subtract 4 √ − 4 √ = 0 Take roots 328 2− 2 = 0 Subtract 0 = 0 True!It works x = 4 Our Solution When there is more than one square root in the problem, after i solating one root and squaring both sides we may still have a root remaining in t he problem. In this case we will again isolate the term with the second root a nd square both sides. When isolating, we will isolate the termwith the square root. This means the square root can be multiplied by a number after isolating .

Example 447. 2x + 1 √ − x √ = 1 Even index !We will have to check answers + x √ + x √ Isolate rst root by adding x √ to both sides 2 x + 1 √ = x √ + 1 Square both sides ( 2 x+ 1 √ ) 2 = ( x √ + 1) 2 Evaluate exponents ,recall (a + b)2 = a2 + 2 a b+b2 2 x + 1 = x+ 2 x √ + 1 Isolate the term with the root − x− 1− x −1 Subtract xand 1from both sides x = 2 x √ Square both sides ( x )2 = (2 x √ ) 2 Evaluate exponents x 2 = 4 x Make equation equal zero − 4x − 4x Subtract xfrom both sides x 2 − 4x = 0 Factor x (x − 4) = 0 Set each factor equal to zero x = 0 orx− 4 = 0 Solve + 4 + 4 Add 4to both sides of second equation x = 0 orx= 4 Need to check answers in original 2(0) + 1 p − (0) p = 1 Check x= 0 rst 1 √ − 0 √ = 1 Take roots 1 − 0 = 1 Subtract 1 = 1 True!It works 2(4) + 1 p − (4) p = 1 Check x= 4 8 + 1 √ − 4 √ = 1 Add 9 √ − 4 √ = 1 Take roots 3 − 2 = 1 Subtract 1 = 1 True!It works 329 x= 0 or4 Our Solution Example 448. 3x + 9 √ − x+ 4 √ = − 1 Even index !We will have to check answers + x+ 4 √ + x+ 4 √ Isolate the rst root by adding x+ 4 √ 3 x + 9 √ = x+ 4 √ − 1 Square both sides ( 3 x+ 9 √ ) 2 = ( x+ 4 √ − 1)2 Evaluate exponents 3 x + 9 = x+ 4 −2 x+ 4 √ + 1 Combine like terms 3 x + 9 = x+ 5 −2 x+ 4 √ Isolate the term with radical − x− 5− x− 5 Subtract xand 5from both sides 2 x + 4 = −2 x+ 4 √ Square both sides (2 x+ 4) 2 = ( −2 x+ 4 √ ) 2 Evaluate exponents 4 x 2 + 16 x+ 16 = 4( x+ 4) Distribute 4 x 2 + 16 x+ 16 = 4 x+ 16 Make equation equal zero − 4x − 16 −4x − 16 Subtract 4x and 16 from both sides 4 x 2 + 12 x= 0 Factor 4 x (x + 3) = 0 Set each factor equal to zero 4 x = 0 orx+ 3 = 0 Solve 4 4 − 3− 3 x= 0 orx= − 3 Check solutions in original 3(0) + 9 p − (0) + 4 p = − 1 Check x= 0 rst 9 √ − 4 √ = − 1 Take roots 3 − 2 = −1 Subtract 1 = −1 False ,extraneous solution 3( −3) + 9 p − (− 3) + 4 p = − 1 Check x= − 3 − 9 + 9 √ − (− 3) + 4 p = − 1 Add 0 √ − 1 √ = − 1 Take roots 0 − 1 = −1 Subtract − 1 = −1 True !It works x = − 3 Our Solution 330 9.1 Practice - Solving with Radicals Solve.

1) 2x + 3 √ − 3 = 0 3) 6x − 5 √ − x= 0 5) 3 + x= 6 x+ 13 √ 7) 3− 3x √ − 1 = 2 x 9) 4x + 5 √ − x+ 4 √ = 2 11) 2x + 4 √ − x+ 3 √ = 1 13) 2x + 6 √ − x+ 4 √ = 1 15) 6− 2x √ − 2x + 3 √ = 3 2) 5x + 1 √ − 4 = 0 4) x+ 2 √ − x √ = 2 6) x− 1 = 7 −x √ 8) 2x + 2 √ = 3 + 2 x− 1 √ 10) 3x + 4 √ − x+ 2 √ = 2 12) 7x + 2 √ − 3x + 6 √ = 6 14) 4x − 3 √ − 3x + 1 √ = 1 16) 2− 3x √ − 3x + 7 √ = 3 331 9.2Quadratics - Solving with Exponents Ob jective: Solve equations with exponents using the odd roo t property and the even root property.

Another type of equation we can solve is one with exponents. A s you might expect we can clear exponents by using roots. This is done wit h very few unex- pected results when the exponent is odd. We solve these probl ems very straight forward using the odd root property Odd Root Property :if an = b , then a= b n √ when nis odd Example 449. x5 = 32 Use odd root property x 5 5 √ = 32 5 √ Simplify roots x = 2 Our Solution However, when the exponent is even we will have two results fr om taking an even root of both sides. One will be positive and one will be negati ve. This is because both 32 = 9 and (− 3)2 = 9 . so when solving x2 = 9 we will have two solutions, one positive and one negative: x= 3 and −3 Even Root Property :if an = b , then a= ± b n √ when nis even Example 450. x4 = 16 Use even root property (± ) 332 x4 4 √ = ± 16 4 √ Simplify roots x = ± 2 Our Solution World View Note: In 1545, French Mathematicain Gerolamo Cardano pub- lished his book The Great Art, or the Rules of Algebra which included the solu- tion of an equation with a fourth power, but it was considered absurd by many to take a quantity to the fourth power because there are only thr ee dimensions!

Example 451. (2x+ 4) 2 = 36 Use even root property (± ) (2 x+ 4) 2 p = ± 36 √ Simplify roots 2 x + 4 = ±6 To avoid sign errors we need two equations 2 x + 4 = 6 or2x + 4 = −6 One equation for +,one equation for − − 4− 4 − 4 −4 Subtract 4from both sides 2 x = 2 or2x = − 10 Divide both sides by 2 2 2 2 2 x = 1 orx= − 5 Our Solutions In the previous example we needed two equations to simplify b ecause when we took the root, our solutions were two rational numbers, 6and −6. If the roots did not simplify to rational numbers we can keep the ±in the equation.

Example 452. (6x− 9)2 = 45 Use even root property (± ) (6 x− 9)2 p = ± 45 √ Simplify roots 6 x − 9 = ±3 5√ Use one equation because root did not simplify to rational + 9 + 9 Add 9to both sides 6 x = 9 ±3 5√ Divide both sides by 6 6 6 x = 9 ± 3 5√ 6 Simplify ,divide each term by 3 x = 3 ± 5 √ 2 Our Solution 333 When solving with exponents, it is important to rst isolatethe part with the exponent before taking any roots.

Example 453.

(x + 4) 3 − 6 = 119 Isolate part with exponent + 6 + 6 ( x + 4) 3 = 125 Use odd root property ( x + 4) 3 3 p = 125 √ Simplify roots x + 4 = 5 Solve − 4− 4 Subtract 4from both sides x = 1 Our Solution Example 454. (6x+ 1) 2 + 6 = 10 Isolate part with exponent − 6− 6 Subtract 6from both sides (6 x+ 1) 2 = 4 Use even root property (± ) (6 x+ 1) 2 p = ± 4 √ Simplify roots 6 x + 1 = ±2 To avoid sign errors ,we need two equations 6 x + 1 = 2 or6x + 1 = −2 Solve each equation − 1− 1 −1 −1 Subtract 1from both sides 6 x = 1 or6x = − 3 Divide both sides by 6 6 6 6 6 x = 1 6 or x= − 1 2 Our Solution When our exponents are a fraction we will need to rst convert the fractional exponent into a radical expression to solve. Recall that am n = ( a n √ ) m . Once we have done this we can clear the exponent using either the even (± ) or odd root property. Then we can clear the radical by raising both sides to an exponent (remember to check answers if the index is even).

Example 455.

(4x+ 1) 2 5 = 9 Rewrite as aradical expression ( 4 x+ 1 5 √ ) 2 = 9 Clear exponent rst with even root property (± ) ( 4 x+ 1 5 √ ) 2 q = ± 9 √ Simplify roots 334 4x + 1 5 √ = ± 3 Clear radical by raising both sides to 5th power ( 4 x+ 1 5 √ ) 5 = ( ±3)5 Simplify exponents 4 x + 1 = ±243 Solve ,need 2equations !

4 x + 1 = 243 or 4x + 1 = −243 − 1 −1 −1 −1 Subtract 1from both sides 4 x = 242 or 4x = − 244 Divide both sides by 4 4 4 4 4 x = 121 2 ,− 61 Our Solution Example 456. (3x− 2)3 4 = 64 Rewrite as radical expression ( 3 x− 2 4 √ ) 3 = 64 Clear exponent rst with odd root property ( 3 x− 2 4 √ ) 3 3 q = 64 3 √ Simplify roots 3 x − 2 4 √ = 4 Even Index !Check answers .

( 3 x− 2 4 √ ) 4 = 4 4 Raise both sides to 4th power 3 x − 2 = 256 Solve + 2 + 2 Add 2to both sides 3 x = 258 Divide both sides by 3 3 3 x = 86 Need to check answer in radical form of problem ( 3( 86)− 2 4 p ) 3 = 64 Multiply ( 258 −2 4 √ ) 3 = 64 Subtract ( 256 4 √ ) 3 = 64 Evaluate root 4 3 = 64 Evaluate exponent 64 =64 True !It works x = 86 Our Solution With rational exponents it is very helpful to convert to radi cal form to be able to see if we need a ±because we used the even root property, or to see if we need to check our answer because there was an even root in the probl em. When checking we will usually want to check in the radical form as i t will be easier to evaluate.

335 9.2 Practice - Solving with Exponents Solve.

1) x2 = 75 3) x2 + 5 = 13 5) 3x 2 + 1 = 73 7) (x + 2) 5 = − 243 9) (2x+ 5) 3 − 6 = 21 11) (x − 1)2 3 = 16 13) (2−x)3 2 = 27 15) (2x− 3) 2 3 = 4 17) (x + 1 2 )− 2 3 = 4 19) (x − 1)− 5 2 = 32 21) (3x− 2) 4 5 = 16 23) (4x+ 2) 3 5 = − 8 2) x3 = − 8 4) 4x 3 − 2 = 106 6) (x − 4)2 = 49 8) (5x+ 1) 4 = 16 10) (2x+ 1) 2 + 3 = 21 12) (x − 1)3 2 = 8 14) (2x+ 3) 4 3 = 16 16) (x + 3) − 1 3 = 4 18) (x − 1)− 5 3 = 32 20) (x + 3) 3 2 = − 8 22) (2x+ 3) 3 2 = 27 24) (3−2x )4 3 = − 81 336 9.3Quadratics - Complete the Square Ob jective: Solve quadratic equations by completing the squ are.

When solving quadratic equations in the past we have used fac toring to solve for our variable. This is exactly what is done in the next example .

Example 457. x2 + 5 x+ 6 = 0 Factor ( x + 3)( x+ 2) = 0 Set each factor equal to zero x + 3 = 0 orx+ 2 = 0 Solve each equation − 3− 3 − 2− 2 x= − 3 or x= − 2 Our Solutions However, the problem with factoring is all equations cannot be factored. Consider the following equation: x2 − 2x − 7 = 0 . The equation cannot be factored, however there are two solutions to this equation, 1 + 2 2√ and 1− 2 2√ . To nd these two solutions we will use a method known as completing the square . When completing the square we will change the quadratic into a perfect square which can easily be solved with the square root property. The next example revie ws the square root property.

Example 458.

(x + 5) 2 = 18 Square root of both sides ( x + 5) 2 p = ± 18 √ Simplify each radical x + 5 = ±3 2√ Subtract 5from both sides − 5 −5 x= − 5± 3 2√ Our Solution 337 To complete the square, or make our problem into the form of the previous example, we will be searching for the third term in a trinomia l. If a quadratic is of the form x2 + b x +c, and a perfect square, the third term, c, can be easily found by the formula 1 2 b 2 . This is shown in the following examples, where we nd the number that completes the square and then factor the p erfect square.

Example 459.

x2 + 8 x+ c c = 1 2 b 2 and our b= 8 1 2 8 2 = 4 2 = 16 The third term to complete the square is 16 x 2 + 8 x+ 16 Our equation as aperfect square ,factor ( x + 4) 2 Our Solution Example 460. x2 − 7x + c c = 1 2 b 2 and our b= 7 1 2 7 2 = 7 2 2 = 49 4 The third term to complete the square is 49 4 x 2 − 11 x+ 49 4 Our equation as aperfect square ,factor x− 7 2 2 Our Solution Example 461. x2 + 5 3 x + c c = 1 2 b 2 and our b= 8 1 2 5 3 2 = 5 6 2 = 25 36 The third term to complete the square is 25 36 338 x2 + 5 3 x + 25 36 Our equation as aperfect square ,factor x+ 5 6 2 Our Solution The process in the previous examples, combined with the even root property, is used to solve quadratic equations by completing the square. The following ve steps describe the process used to complete the square, alon g with an example to demonstrate each step.

Problem 3 x 2 + 18 x− 6 = 0 1. Separate constant term from variables + 6 + 6 3x 2 + 18 x = 6 2. Divide each term by a 3 3x2 + 18 3 x =6 3 x 2 + 6 x = 2 3. Find value to complete the square: 1 2 b 2 1 2 6 2 = 3 2 = 9 4. Add to both sides of equation x 2 + 6 x = 2 + 9 + 9 x 2 + 6 x+ 9 = 11 5. Factor (x + 3) 2 = 11 Solve by even root property (x + 3) 2 p = ± 11 √ x + 3 = ±11 √ − 3 −3 x= − 3± 11 √ World View Note:

The Chinese in 200 BC were the rst known culture group to use a method similar to completing the square, but their me thod was only used to calculate positive roots.

The advantage of this method is it can be used to solve any quad ratic equation.

The following examples show how completing the square can gi ve us rational solu- tions, irrational solutions, and even complex solutions.

Example 462.

2x 2 + 20 x+ 48 = 0 Separate constant term from varaibles 339 −48 −48 Subtract 24 2 x 2 + 20 x =− 48 Divide by aor 2 2 2 2 x 2 + 10 x =− 24 Find number to complete the square : 1 2 b 2 1 2 10 2 = 5 2 = 25 Add 25 to both sides of the equation x 2 + 10 x =− 24 + 25 +25 x2 + 10 x+ 25 = 1 Factor ( x + 5) 2 = 1 Solve with even root property ( x + 5) 2 p = ± 1 √ Simplify roots x + 5 = ±1 Subtract 5from both sides − 5− 5 x= − 5± 1 Evaluate x = − 4 or −6 Our Solution Example 463. x2 − 3x − 2 = 0 Separate constant from variables + 2 + 2 Add 2to both sides x 2 − 3x = 2 Noa, nd number to complete the square 1 2 b 2 1 2 3 2 = 3 2 2 = 9 4 Add 9 4 to both sides , 2 1 4 4 + 9 4 =8 4 + 9 4 =17 4 Need common denominator (4)on right x 2 − 3x + 9 4 =8 4 + 9 4 =17 4 Factor x− 3 2 2 = 17 4 Solve using the even root property x− 3 2 2 s = ± 17 4 r Simplify roots x − 3 2 =± 17 √ 2 Add 3 2 to both sides , 340 +3 2 + 3 2 we already have acommon denominator x = 3 ± 17 √ 2 Our Solution Example 464. 3x 2 = 2 x− 7 Separate the constant from the variables − 2x − 2x Subtract 2x from both sides 3 x 2 − 2x = − 7 Divide each term by aor 3 3 3 3 x 2 − 2 3 x =− 7 3 Find the number to complete the square 1 2 b 2 1 2 2 3 2 = 1 3 2 = 1 9 Add to both sides , − 7 3 3 3 + 1 9 = − 21 3 + 1 9 =− 20 9 get common denominator on right x 2 − 2 3 x + 1 3 = − 20 9 Factor x− 1 3 2 = − 20 9 Solve using the even root property x− 1 3 2 s = ± − 20 9 r Simplify roots x − 1 3 =± 2i 5 √ 3 Add 1 3 to both sides , + 1 3 +1 3 Already have common denominator x = 1 ± 2i 5 √ 3 Our Solution As several of the examples have shown, when solving by comple ting the square we will often need to use fractions and be comfortable nding co mmon denominators and adding fractions together. Once we get comfortable solv ing by completing the square and using the ve steps, any quadratic equation can be easily solved.

341 9.3 Practice - Complete the Square Find the value that completes the square and then rewrite as a perfect square.

1) x2 − 30 x+ __ 3) m2 − 36 m +__ 5) x2 − 15 x+ __ 7) y2 − y+ __ 2) a2 − 24 a+ __ 4) x2 − 34 x+ __ 6) r2 − 1 9 r + __ 8) p2 − 17 p+ __ Solve each equation by completing the square.

9) x2 − 16 x+ 55 = 0 11) v2 − 8v + 45 = 0 13) 6x 2 + 12 x+ 63 = 0 15) 5k 2 − 10 k+ 48 = 0 17) x2 + 10 x− 57 = 4 19) n2 − 16 n+ 67 = 4 21) 2x 2 + 4 x+ 38 =− 6 23) 8b2 + 16 b− 37 = 5 25) x2 = − 10 x− 29 27) n2 = − 21 +10 n 29) 3k 2 + 9 = 6 k 31) 2x 2 + 63 = 8 x 33) p2 − 8p = − 55 35) 7n 2 − n+ 7 = 7 n+ 6 n2 37) 13 b2 + 15 b+ 44 =− 5 + 7 b2 + 3 b 39) 5x 2 + 5 x= − 31 −5x 41) v2 + 5 v+ 28 = 0 43) 7x 2 − 6x + 40 = 0 45) k2 − 7k + 50 = 3 47) 5x 2 + 8 x− 40 = 8 49) m2 = − 15 + 9 m 51) 8r2 + 10 r= − 55 53) 5n 2 − 8n + 60 =− 3n + 6 + 4 n2 55) −2x 2 + 3 x− 5 = −4x 2 10) n2 − 8n − 12 = 0 12) b2 + 2 b+ 43 = 0 14) 3x 2 − 6x + 47 = 0 16) 8a 2 + 16 a− 1 = 0 18) p2 − 16 p− 52 = 0 20) m2 − 8m −3 = 6 22) 6r2 + 12 r− 24 =− 6 24) 6n 2 − 12 n− 14 = 4 26) v2 = 14 v+ 36 28) a2 − 56 =− 10 a 30) 5x 2 = − 26 +10 x 32) 5n 2 = − 10 n+ 15 34) x2 + 8 x+ 15 = 8 36) n2 + 4 n= 12 38) −3r2 + 12 r+ 49 =− 6r2 40) 8n 2 + 16 n= 64 42) b2 + 7 b− 33 = 0 44) 4x 2 + 4 x+ 25 = 0 46) a2 − 5a + 25 = 3 48) 2p2 − p+ 56 =− 8 50) n2 − n= − 41 52) 3x 2 − 11 x= − 18 54) 4b2 − 15 b+ 56 = 3 b2 56) 10 v2 − 15 v= 27 + 4 v2 − 6v 342 9.4Quadratics - Quadratic Formula Ob jective: Solve quadratic equations by using the quadrati c formula.

The general from of a quadratic is a x2 + b x +c= 0 . We will now solve this for- mula for xby completing the square Example 465. a x2 + b c +c= 0 Separate constant from variables − c− c Subtract cfrom both sides a x 2 + b x =− c Divide each term by a a a a x 2 + b a x =− c a Find the number that completes the square 1 2 b a 2 = b 2 a 2 = b 2 4 a 2 Add to both sides , b 2 4 a 2 − c a 4a 4 a = b 2 4 a 2 − 4 a c 4 a 2 = b 2 − 4a c 4 a 2 Get common denominator on right x 2 + b a x + b 2 4 a 2 = b 2 4 a 2 − 4 ac 4 a 2 = b 2 − 4a c 4 a 2 Factor x+ b 2 a 2 = b 2 − 4a c 4 a 2 Solve using the even root property x+ b 2 a 2 s = ± b 2 − 4a c 4 a 2 r Simplify roots x + b 2 a = ± b2 − 4a c √ 2 a Subtract b 2 a from both sides x = − b± b2 − 4a c √ 2 a Our Solution This solution is a very important one to us. As we solved a gene ral equation by completing the square, we can use this formula to solve any qu adratic equation.

Once we identify what a , b,andcare in the quadratic, we can substitute those 343 values intox= − b± b2 − 4ac p 2 a and we will get our two solutions. This formula is known as the quadratic fromula Quadratic Formula :if a x 2 + b x +c= 0 then x= − b± b2 − 4a c p 2 a World View Note: Indian mathematician Brahmagupta gave the rst explicit formula for solving quadratics in 628. However, at that time mathematics was not done with variables and symbols, so the formula he gave was, “ To the absolute number multiplied by four times the square, add the square of the middle term; the square root of the same, less the middle term, being divid ed by twice the square is the value.” This would translate to 4ac +b2 p − b 2 a as the solution to the equation a x2 + b x =c.

We can use the quadratic formula to solve any quadratic, this is shown in the fol- lowing examples.

Example 466.

x2 + 3 x+ 2 = 0 a= 1 , b= 3 , c= 2 ,use quadratic formula x = − 3± 32 − 4(1)(2) p 2(1) Evaluate exponent and multiplication x = − 3± 9− 8 √ 2 Evaluate subtraction under root x = − 3± 1 √ 2 Evaluate root x = − 3± 1 2 Evaluate ±to get two answers x = − 2 2 or − 4 2 Simplify fractions x = − 1 or −2 Our Solution As we are solving using the quadratic formula, it is importan t to remember the equation must st be equal to zero.

Example 467.

25x2 = 30 x+ 11 First set equal to zero − 30 x− 11 −30 x− 11 Subtract 30 xand 11 from both sides 25 x2 − 30 x− 11 = 0 a= 25 , b =− 30 , c =− 11 ,use quadratic formula x = 30 ± (− 30 )2 − 4(25 )(− 11 ) p 2( 25 ) Evaluate exponent and multiplication 344 x= 30 ± 900+1100 √ 50 Evaluate addition inside root x = 30 ± 2000 √ 50 Simplify root x = 30 ±20 5 √ 50 Reduce fraction by dividing each term by 10 x = 3 ± 2 5√ 5 Our Solution Example 468. 3x 2 + 4 x+ 8 = 2 x2 + 6 x− 5 First set equation equal to zero − 2x 2 − 6x + 5 −2x 2 − 6x + 5 Subtract 2x 2 and 6x and add 5 x 2 − 2x + 13 = 0 a= 1 , b=− 2, c =13 ,use quadratic formula x = 2 ± (− 2)2 − 4(1)( 13) p 2(1) Evaluate exponent and multiplication x = 2 ± 4− 52 √ 2 Evaluate subtraction inside root x = 2 ± − 48 √ 2 Simplify root x = 2 ± 4i 3 √ 2 Reduce fraction by dividing each term by 2 x = 1 ±2i 3 √ Our Solution When we use the quadratic formula we don’t necessarily get tw o unique answers.

We can end up with only one solution if the square root simpli es to zero.

Example 469. 4x 2 − 12 x+ 9 = 0 a= 4 , b=− 12 , c = 9 ,use quadratic formula x = 12 ± (− 12 )2 − 4(4)(9) p 2(4) Evaluate exponents and multiplication x = 12 ± 144−144 √ 8 Evaluate subtraction inside root x = 12 ± 0 √ 8 Evaluate root x = 12 ±0 8 Evaluate ± x = 12 8 Reduce fraction x = 3 2 Our Solution 345 If a term is missing from the quadratic, we can still solve with the quadratic for- mula, we simply use zero for that term. The order is important , so if the term with xis missing, we have b= 0 , if the constant term is missing, we have c= 0 .

Example 470. 3x 2 + 7 = 0 a= 3 , b= 0 ( missing term ), c = 7 x = − 0± 02 − 4(3)(7) p 2(3) Evaluate exponnets and multiplication ,zeros not needed x = ± − 84 √ 6 Simplify root x = ± 2i 21 √ 6 Reduce ,dividing by 2 x = ± i 21 √ 3 Our Solution We have covered three di erent methods to use to solve a quadr atic: factoring, complete the square, and the quadratic formula. It is import ant to be familiar with all three as each has its advantage to solving quadratic s. The following table walks through a suggested process to decide which method wou ld be best to use for solving a problem.

1. If it can easily factor, solve by factoring x 2 − 5x + 6 = 0 ( x − 2)( x− 3) = 0 x = 2 orx= 3 2. If a= 1 and bis even, complete the square x2 + 2 x= 4 1 2 2 2 = 1 2 = 1 x 2 + 2 x+ 1 = 5 ( x + 1) 2 = 5 x + 1 = ±5 √ x = − 1± 5 √ 3. Otherwise, solve by the quadratic formula x 2 − 3x + 4 = 0 x = 3 ± (− 3)2 − 4(1)(4) p 2(1) x = 3 ± i 7 √ 2 The above table is mearly a suggestion for deciding how to sol ve a quadtratic.

Remember completing the square and quadratic formula will a lways work to solve any quadratic. Factoring only woks if the equation can be fac tored.

346 9.4 Practice - Quadratic Formula Solve each equation with the quadratic formula.

1) 4a 2 + 6 = 0 3) 2x 2 − 8x − 2 = 0 5) 2m 2 − 3 = 0 7) 3r2 − 2r − 1 = 0 9) 4n 2 − 36 = 0 11) v2 − 4v − 5 = −8 13) 2a 2 + 3 a+ 14 = 6 15) 3k 2 + 3 k− 4 = 7 17) 7x 2 + 3 x− 16 =− 2 19) 2p2 + 6 p− 16 = 4 21) 3n 2 + 3 n= − 3 23) 2x 2 = − 7x + 49 25) 5x 2 = 7 x+ 7 27) 8n 2 = − 3n − 8 29) 2x 2 + 5 x= − 3 31) 4a 2 − 64 = 0 33) 4p2 + 5 p− 36 = 3 p2 35) −5n 2 − 3n − 52 = 2 −7n 2 37) 7r2 − 12 =− 3r 39) 2n 2 − 9 = 4 2) 3k 2 + 2 = 0 4) 6n 2 − 1 = 0 6) 5p2 + 2 p+ 6 = 0 8) 2x 2 − 2x − 15 = 0 10) 3b2 + 6 = 0 12) 2x 2 + 4 x+ 12 = 8 14) 6n 2 − 3n + 3 = −4 16) 4x 2 − 14 =− 2 18) 4n 2 + 5 n= 7 20) m2 + 4 m−48 =− 3 22) 3b2 − 3 = 8 b 24) 3r2 + 4 = −6r 26) 6a 2 = − 5a + 13 28) 6v 2 = 4 + 6 v 30) x2 = 8 32) 2k 2 + 6 k− 16 = 2 k 34) 12 x2 + x+ 7 = 5 x2 + 5 x 36) 7m 2 − 6m + 6 = −m 38) 3x 2 − 3 = x2 40) 6b2 = b2 + 7 −b 347 9.5Quadratics - Build Quadratics From Roots Ob jective: Find a quadratic equation that has given roots us ing reverse factoring and reverse completing the square.

Up to this point we have found the solutions to quadratics by a method such as factoring or completing the square. Here we will take our sol utions and work backwards to nd what quadratic goes with the solutions.

We will start with rational solutions. If we have rational so lutions we can use fac- toring in reverse, we will set each solution equal to xand then make the equation equal to zero by adding or subtracting. Once we have done this our expressions will become the factors of the quadratic.

Example 471.

The solutions are 4and −2 Set each solution equal to x x = 4 orx= − 2 Make each equation equal zero − 4− 4 + 2 + 2 Subtract 4from rst ,add 2to second x − 4 = 0 orx+ 2 = 0 These expressions are the factors ( x − 4)( x+ 2) = 0 FOIL x 2 + 2 x− 4x − 8 Combine like terms x 2 − 2x − 8 = 0 Our Solution If one or both of the solutions are fractions we will clear the fractions by multi- plying by the denominators.

Example 472.

The solution are 2 3 and3 4 Set each solution equal to x x = 2 3 or x= 3 4 Clear fractions by multiplying by denominators 3 x = 2 or4x = 3 Make each equation equal zero − 2− 2 −3− 3 Subtract 2from the rst ,subtract 3from the second 3 x − 2 = 0 or4x − 3 = 0 These expressions are the factors (3 x− 2)(4 x− 3) = 0 FOIL 12 x2 − 9x − 8x + 6 = 0 Combine like terms 348 12x2 − 17 x+ 6 = 0 Our Solution If the solutions have radicals (or complex numbers) then we c annot use reverse factoring. In these cases we will use reverse completing the square. When there are radicals the solutions will always come in pairs, one wit h a plus, one with a minus, that can be combined into “one” solution using ±. We will then set this solution equal to xand square both sides. This will clear the radical from our problem.

Example 473.

The solutions are 3 √ and − 3 √ Write as ′′ one ′′ expression equal to x x = ± 3 √ Square both sides x 2 = 3 Make equal to zero − 3− 3 Subtract 3from both sides x 2 − 3 = 0 Our Solution We may have to isolate the term with the square root (with plus or minus) by adding or subtracting. With these problems, remember to squ are a binomial we use the formula (a + b)2 = a2 + 2 a b+b2 Example 474. The solutions are 2− 5 2√ and 2 + 5 2 √ Write as ′′ one ′′ expression equal to x x = 2 ±5 2√ Isolate the square root term − 2− 2 Subtract 2from both sides x − 2 = ±5 2√ Square both sides x 2 − 4x + 4 = 25 2 x 2 − 4x + 4 = 50 Make equal to zero − 50 −50 Subtract 50 x 2 − 4x − 46 = 0 Our Solution World View Note: Before the quadratic formula, before completing the square , before factoring, quadratics were solved geometrically by the Greeks as early as 300 BC! In 1079 Omar Khayyam, a Persian mathematician solved cubic equations geometrically!

If the solution is a fraction we will clear it just as before by multiplying by the denominator.

349 Example 475.The solutions are 2 + 3√ 4 and 2 − 3 √ 4 Write as ′′ one ′′ expresion equal to x x = 2 ± 3 √ 4 Clear fraction by multiplying by 4 4 x = 2 ± 3 √ Isolate the square root term − 2− 2 Subtract 2from both sides 4 x − 2 = ± 3 √ Square both sides 16 x2 − 16 x+ 4 = 3 Make equal to zero − 3− 3 Subtract 3 16 x2 − 16 x+ 1 = 0 Our Solution The process used for complex solutions is identical to the pr ocess used for radi- cals.

Example 476.

The solutions are 4− 5i and 4 + 5 iWrite as ′′ one ′′ expression equal to x x = 4 ±5i Isolate the iterm − 4− 4 Subtract 4from both sides x − 4 = ±5i Square both sides x 2 − 8x + 16 =25 i2 i2 = − 1 x 2 − 8x + 16 =− 25 Make equal to zero + 25 +25 Add 25 to both sides x 2 − 8x + 41 = 0 Our Solution Example 477. The solutions are 3 − 5i 2 and 3 + 5 i 2 Write as ′′ one ′′ expression equal to x x = 3 ± 5i 2 Clear fraction by multiplying by denominator 2 x = 3 ±5i Isolate the iterm − 3− 3 Subtract 3from both sides 2 x − 3 = ±5i Square both sides 4 x 2 − 12 x+ 9 = 5 i2 i2 = − 1 4 x 2 − 12 x+ 9 = −25 Make equal to zero + 25 +25 Add 25 to both sides 4 x 2 − 12 x+ 34 = 0 Our Solution 350 9.5 Practice - Build Quadratics from Roots From each problem, nd a quadratic equation with those numbe rs as its solutions.

1) 2, 5 3) 20, 2 5) 4, 4 7) 0, 0 9) −4,11 11) 3 4 ,1 4 13) 1 2 ,1 3 15) 3 7 , 4 17) −1 3 ,5 6 19) −6, 1 9 21) ±5 23) ±1 5 25) ±11 √ 27) ±3 √ 4 29) ±i 13 √ 31) 2± 6 √ 33) 1± 3i 35) 6± i 3 √ 37) − 1± 6 √ 2 39) 6 ± i 2 √ 8 2) 3, 6 4) 13, 1 6) 0, 9 8) −2,− 5 10) 3, −1 12) 5 8 ,5 7 14) 1 2 ,2 3 16) 2, 2 9 18) 5 3 , − 1 2 20) −2 5 , 0 22) ±1 24) ±7 √ 26) ±2 3√ 28) ±11 i 30) ±5i 2 √ 32) −3± 2 √ 34) −2± 4i 36) −9± i 5 √ 38) 2 ± 5i 3 40) − 2± i 15 √ 2 351 9.6Quadratics - Quadratic in Form Ob jective: Solve equations that are quadratic in form by sub stitution to create a quadratic equation.

We have seen three di erent ways to solve quadratics: factor ing, completing the square, and the quadratic formula. A quadratic is any equati on of the form0 = a x 2 + b x +c, however, we can use the skills learned to solve quadratics t o solve problems with higher (or sometimes lower) powers if the equa tion is in what is called quadratic form.

Quadratic Form : 0 =a xm + b x n + cwhere m= 2 n An equation is in quadratic form if one of the exponents on a va riable is double the exponent on the same variable somewhere else in the equat ion. If this is the case we can create a new variable, set it equal to the variable with smallest expo- nent. When we substitute this into the equation we will have a quadratic equa- tion we can solve.

World View Note: Arab mathematicians around the year 1000 were the rst to use this method!

Example 478.

x4 − 13 x2 + 36 = 0 Quadratic form ,one exponent ,4 ,double the other ,2 y = x2 New variable equal to the variable with smaller exponent y 2 = x4 Square both sides y 2 − 13 y+ 36 = 0 Substitute yfor x2 and y2 for x4 ( y − 9)( y− 4) = 0 Solve.We can solve this equation by factoring y − 9 = 0 ory− 4 = 0 Set each factor equal to zero + 9 + 9 + 4 + 4 Solve each equation y = 9 ory= 4 Solutions for y ,need x.We will use y= x2 equation 9 = x2 or 4 = x2 Substitute values for y ± 9 √ = x2 √ or ± 4 √ = x2 √ Solve using the even root property ,simplify roots x = ± 3,± 2 Our Solutions When we have higher powers of our variable, we could end up wit h many more solutions. The previous equation had four unique solutions .

352 Example 479.a− 2 − a− 1 − 6 = 0 Quadratic form ,one exponent ,− 2,is double the other ,− 1 b = a− 1 Make anew variable equal to the variable with lowest exponent b 2 = a− 2 Square both sides b 2 − b− 6 = 0 Substitute b2 for a− 2 and bfor a− 1 ( b − 3)( b+ 2) = 0 Solve.We will solve by factoring b − 3 = 0 orb+ 2 = 0 Set each factor equal to zero + 3 + 3 −2− 2 Solve each equation b = 3 orb= − 2 Solutions for b ,still need a,substitute into b= a− 1 3 = a− 1 or −2 = a− 1 Raise both sides to −1power 3 − 1 = a or (− 2)− 1 = a Simplify negative exponents a = 1 3 , − 1 2 Our Solution Just as with regular quadratics, these problems will not alw ays have rational solu- tions. We also can have irrational or complex solutions to ou r equations.

Example 480. 2x 4 + x2 = 6 Make equation equal to zero − 6− 6 Subtract 6from both sides 2 x 4 + x2 − 6 = 0 Quadratic form ,one exponent ,4 ,double the other ,2 y = x2 New variable equal variable with smallest exponent y 2 = x4 Square both sides 2 y2 + y− 6 = 0 Solve.We will factor this equation (2 y− 3)( y+ 2) = 0 Set each factor equal to zero 2 y − 3 = 0 ory+ 2 = 0 Solve each equation + 3 + 3 − 2− 2 2y = 3 ory= − 2 2 2 y = 3 2 or y= − 2 We have y ,still need x.Substitute into y= x2 3 2 = x2 or −2 = x2 Square root of each side ± 3 2 r = x2 √ or ± − 2 √ = x2 √ Simplify each root ,rationalize denominator x = ± 6 √ 2 , ± i 2 √ Our Solution 353 When we create a new variable for our substitution, it won’t always be equal to just another variable. We can make our substitution variabl e equal to an expres- sion as shown in the next example.

Example 481.

3(x− 7)2 − 2(x− 7) + 5 = 0 Quadratic form y = x− 7 De ne new variable y 2 = ( x− 7)2 Square both sides 3 y2 − 2y + 5 = 0 Substitute values into original (3 y− 5)( y+ 1) = 0 Factor 3 y − 5 = 0 ory+ 1 = 0 Set each factor equal to zero + 5 + 5 − 1− 1 Solve each equation 3 y = 5 ory= − 1 3 3 y = 5 3 or y= − 1 We have y ,we still need x.

5 3 = x− 7 or −1 = x− 7 Substitute into y= x− 7 + 21 3 + 7 + 7 + 7 Add 7.Use common denominator as needed x = 26 3 , 6 Our Solution Example 482. (x 2 − 6x )2 = 7( x2 − 6x )− 12 Make equation equal zero − 7(x2 − 6x ) + 12−7(x2 − 6x ) + 12 Move all terms to left ( x 2 − 6x )2 − 7(x2 − 6x ) + 12= 0 Quadratic form y = x2 − 6x Make new variable y 2 = ( x2 − 6x )2 Square both sides y 2 − 7y + 12 = 0 Substitute into original equation ( y − 3)( y− 4) = 0 Solve by factoring y − 3 = 0 ory− 4 = 0 Set each factor equal to zero + 3 + 3 + 4 + 4 Solve each equation y = 3 ory= 4 We havey ,still need x.

3 = x2 − 6x or 4 = x3 − 6x Solve each equation ,complete the square 1 2 6 2 = 3 2 = 9 Add9to both sides of each equation 12 =x2 − 6x + 9 or 13 =x2 − 6x + 9 Factor 354 12= ( x− 3)2 or 13 = (x− 3)2 Use even root property ± 12 √ = ( x− 3)2 p or ± 13 √ = ( x− 3)2 p Simplify roots ± 2 3√ = x− 3 or ± 13 √ = x− 3 Add 3to both sides + 3 + 3 + 3 + 3 x = 3 ±2 3√ , 3 ± 13 √ Our Solution The higher the exponent, the more solution we could have. Thi s is illustrated in the following example, one with six solutions.

Example 483.

x6 − 9x 3 + 8 = 0 Quadratic form ,one exponent ,6 ,double the other ,3 y = x3 New variable equal to variable with lowest exponent y 2 = x6 Square both sides y 2 − 9y + 8 = 0 Substitute y2 for x6 and yfor x3 ( y − 1)( y− 8) = 0 Solve.We will solve by factoring .

y − 1 = 0 ory− 8 = 0 Set each factor equal to zero + 1 + 1 + 8 + 8 Solve each equation y = 1 ory= 8 Solutions for y ,we need x.Substitute into y= x3 x 3 = 1 orx3 = 8 Set each equation equal to zero − 1− 1 −8− 8 x3 − 1 = 0 orx3 − 8 = 0 Factor each equation ,di erence of cubes ( x − 1)( x2 + x+ 1) = 0 First equation factored .Set each factor equal to zero x − 1 = 0 orx2 + x+ 1 = 0 First equation is easy to solve + 1 + 1 x = 1 First solution − 1± 12 − 4(1)(1) p 2(1) = 1 ± i 3 √ 2 Quadratic formula on second factor ( x − 2)( x2 + 2 x+ 4) = 0 Factor the second di erence of cubes x − 2 = 0 orx2 + 2 x+ 4 = 0 Set each factor equal to zero .

+ 2 + 2 First equation is easy to solve x = 2 Our fourth solution − 2± 22 − 4(1)(4) p 2(1) = − 1± i 3 √ Quadratic formula on second factor x = 1 ,2 , 1 ± i 3 √ 2 , − 1± i 3 √ Our nal six solutions 355 9.6 Practice - Quadratic in Form Solve each of the following equations. Some equations will h ave complex roots.

1) x4 − 5x 2 + 4 = 0 3) m4 − 7m 2 − 8 = 0 5) a4 − 50 a2 + 49 = 0 7) x4 − 25 x2 + 144 = 0 9) m4 − 20 m2 + 64 = 0 11) z6 − 216 =19 z3 13) 6z 4 − z2 = 12 15) x2 3 − 35 = 2 x1 3 17) y− 6 + 7 y− 3 = 8 19) x4 − 2x 2 − 3 = 0 21) 2x 4 − 5x 2 + 2 = 0 23) x4 − 9x 2 + 8 = 0 25) 8x 6 − 9x 3 + 1 = 0 27) x8 − 17 x4 + 16 = 0 29) (y + b)2 − 4(y+ b) = 21 31) (y + 2) 2 − 6(y+ 2) = 16 33) (x − 3)2 − 2(x− 3) = 35 35) (r − 1)2 − 8(r− 1) = 20 37) 3(y+ 1) 2 − 14 (y + 1) = 5 39) (3x2 − 2x )2 + 5 = 6(3 x2 − 2x ) 41) 2(3x+ 1) 2 3 − 5(3 x+ 1) 1 3 = 88 43) (x 2 + 2 x)2 − 2(x2 + 2 x) = 3 45) (2x2 − x)2 − 4(2 x2 − x) + 3 = 0 2) y4 − 9y2 + 20 = 0 4) y4 − 29 y2 + 100 = 0 6) b4 − 10 b2 + 9 = 0 8) y4 − 40 y2 + 144 = 0 10) x6 − 35 x3 + 216 = 0 12) y4 − 2y2 = 24 14) x− 2 − x− 1 − 12 = 0 16) 5y− 2 − 20 =21 y− 1 18) x4 − 7x 2 + 12 = 0 20) x4 + 7 x2 + 10 = 0 22) 2x 4 − x2 − 3 = 0 24) x6 − 10 x3 + 16 = 0 26) 8x 6 + 7 x3 − 1 = 0 28) (x − 1)2 − 4(x− 1) = 5 30) (x + 1) 2 + 6( x+ 1) + 9 = 0 32) (m −1)2 − 5(m −1) = 14 34) (a + 1) 2 + 2( a− 1) = 15 36) 2(x− 1)2 − (x − 1) = 3 38) (x 2 − 3)2 − 2(x2 − 3) = 3 40) (x 2 + x+ 3) 2 + 15 = 8( x2 + x+ 3) 42) (x 2 + x)2 − 8(x2 + x) + 12= 0 44) (2x2 + 3 x)2 = 8(2 x2 + 3 x) + 9 46) (3x2 − 4x )2 = 3(3 x2 − 4x ) + 4 356 9.7Quadratics - Rectangles Ob jective: Solve applications of quadratic equations usin g rectangles.

An application of solving quadratic equations comes from th e formula for the area of a rectangle. The area of a rectangle can be calculated by mu ltiplying the width by the length. To solve problems with rectangles we will rst draw a picture to represent the problem and use the picture to set up our equati on.

Example 484.

The length of a rectangle is 3 more than the width. If the area i s 40 square inches, what are the dimensions?

40 xWe do not know the width , x.

x + 3 Length is4more ,or x+ 4 ,and area is 40 .

357 x(x + 3) = 40 Multiply length by width to get area x 2 + 3 x= 40 Distribute − 40 −40 Make equation equal zero x 2 + 3 x− 40 = 0 Factor ( x − 5)( x+ 8) = 0 Set each factor equal to zero x − 5 = 0 orx+ 8 = 0 Solve each equation + 5 + 5 − 8− 8 x= 5 orx= − 8 Our xis awidth ,cannot be negative .

(5) + 3 = 8 Length isx+ 3 ,substitute 5for xto nd length 5 in by 8in Our Solution The above rectangle problem is very simple as there is only on e rectangle involved. When we compare two rectangles, we may have to get a bit more cre- ative.

Example 485.

If each side of a square is increased by 6, the area is multipli ed by 16. Find the side of the original square.

x2 x Square has all sides the same length x Area is found by multiplying length by width 16 x2 x + 6 Each side is increased by 6, x + 6 Area is 16 times original area ( x + 6)( x+ 6) = 16x2 Multiply length by width to get area x 2 + 12 x+ 36 =16 x2 FOIL − 16 x2 −16 x2 Make equation equal zero − 15 x2 + 12 x+ 36 = 0 Divide each term by −1,changes the signs 15 x2 − 12 x− 36 = 0 Solve using the quadratic formula x = 12 ± (− 12 )2 − 4(15 )(− 36 ) p 2( 15 ) Evaluate x = 16 ± 2304 √ 30 x = 16 ±48 30 Can ′ t have anegative solution ,we will only add x = 60 30 = 2 Ourxis the original square 358 2Our Solution Example 486.

The length of a rectangle is 4 ft greater than the width. If eac h dimension is increased by 3, the new area will be 33 square feet larger. Fin d the dimensions of the original rectangle.

x(x + 4) x We don ′ t know width , x ,length is 4more , x+ 4 x + 4 Area is found by multiplying length by width x (x + 4) + 33 x+ 3 Increase each side by 3.

width becomes x+ 3 ,length x+ 4 + 3 = x+ 7 x + 7 Area is 33 more than original , x(x + 4) + 33 ( x + 3)( x+ 7) = x(x + 4) + 33 Set up equation ,length times width is area x 2 + 10 x+ 21 =x2 + 4 x+ 33 Subtract x2 from both sides − x2 −x2 10 x+ 21 = 4 x+ 33 Move variables to one side − 4x −4x Subtract 4x from each side 6 x + 21 =33 Subtract 21 from both sides − 21 −21 6x = 12 Divide both sides by 6 6 6 x = 2 xis the width of the original (2) + 4 = 6 x+ 4 is the length .Substitute 2to nd 2 ft by 6ft Our Solution From one rectangle we can nd two equations. Perimeter is fou nd by adding all the sides of a polygon together. A rectangle has two widths an d two lengths, both the same size. So we can use the equation P= 2 l+ 2 w(twice the length plus twice the width).

Example 487.

The area of a rectangle is 168 cm 2 . The perimeter of the same rectangle is 52 cm.

What are the dimensions of the rectangle?

x We don ′ t know anything about length or width y Use two variables , xand y x y =168 Length times width gives the area .

2 x + 2 y= 52 Also use perimeter formula .

− 2x −2x Solve by substitution ,isolate y 2 y = − 2x + 52 Divide each term by 2 359 2 2 2 y= − x+ 26 Substitute into area equation x (− x+ 26 ) = 168 Distribute − x2 + 26 x= 168 Divide each term by −1,changing all the signs x 2 − 26 x= − 168 Solve by completing the square .

1 2 26 2 = 13 2 = 169 Find number to complete the square : 1 2 b 2 x 2 − 26 x+ 324 = 1 Add 169 to both sides ( x − 13 )2 = 1 Factor x − 13 =± 1 Square root both sides + 13 +13 x= 13 ±1 Evaluate x = 14 or 12 Two options for rst side .

y = − (14 ) + 26=12 Substitute 14 into y= − x+ 26 y = − (12 ) + 26=14 Substitute 12 into y= − x+ 26 Both are the same rectangle ,variables switched !

12 cm by 14 cm Our Solution World View Note: Indian mathematical records from the 9th century demon- strate that their civilization had worked extensivly in geo metry creating religious alters of various shapes including rectangles.

Another type of rectangle problem is what we will call a “fram e problem”. The idea behind a frame problem is that a rectangle, such as a phot ograph, is centered inside another rectangle, such as a frame. In these cases it w ill be important to rememember that the frame extends on all sides of the rectang le. This is shown in the following example.

Example 488.

An 8 in by 12 in picture has a frame of uniform width around it. T he area of the frame is equal to the area of the picture. What is the width of t he frame?

8 12 12 + 2 x Draw picture ,picture if 8by 10 If frame has width x ,on both sides ,we add 2x 8 + 2 x 8 12 =96 Area of the picture ,length times width 2 96 =192 Frame is the same as the picture .Total area is double this .

( 12 + 2 x)(8 + 2 x) = 192 Area of everything ,length times width 96 +24 x+ 16 x+ 4 x2 = 192 FOIL 4 x 2 + 40 x+ 96 =192 Combine like terms − 192 −192 Make equation equal to zero by subtracting 192 4 x 2 + 40 x− 96 = 0 Factor out GCF of 4 360 4(x2 + 10 x− 24 ) = 0 Factor trinomial 4( x− 2)( x+ 12 ) = 0 Set each factor equal to zero x − 2 = 0 orx+ 12 = 0 Solve each equation + 2 + 2 − 12 −12 x= 2 or−12 Can ′ t have negative frame width .

2 inches Our Solution Example 489.

A farmer has a eld that is 400 rods by 200 rods. He is mowing the eld in a spiral pattern, starting from the outside and working in tow ards the center. After an hour of work, 72% of the eld is left uncut. What is the size o f the ring cut around the outside?

400−2x 200 −2x 200 Draw picture ,outside is 200 by 400 If frame has width xon both sides , subtract 2x from each side to get center 400 400 200 =80000 Area of entire eld ,length times width 80000 (0.72 ) =57600 Area of center ,multiply by 28 %as decimal ( 400 −2x )( 200 −2x ) = 57600 Area of center ,length times width 80000 −800 x− 400 x+ 4 x2 = 57600 FOIL 4 x 2 − 1200 x+ 80000 =57600 Combine like terms − 57600 −57600 Make equation equal zero 4 x 2 − 1200 x+ 22400 = 0 Factor out GCF of 4 4( x2 − 300 x+ 5600 ) = 0 Factor trinomial 4( x− 280 )(x− 20 ) = 0 Set each factor equal to zero x − 280 = 0 orx− 20 = 0 Solve each equation + 280 +280 +20 +20 x= 280 or 20 The eld is only 200 rods wide , Can ′ t cut 280 o two sides !

20 rods Our Solution For each of the frame problems above we could have also comple ted the square or use the quadratic formula to solve the trinomials. Remember that completing the square or the quadratic formula always will work when solvin g, however, factoring only works if we can factor the trinomial.

361 9.7 Practice - Rectangles 1) In a landscape plan, a rectangular owerbed is designed to be 4 meters longer than it is wide. If 60 square meters are needed for the plants i n the bed, what should the dimensions of the rectangular bed be?

2) If the side of a square is increased by 5 the area is multipli ed by 4. Find the side of the original square.

3) A rectangular lot is 20 yards longer than it is wide and its a rea is 2400 square yards. Find the dimensions of the lot.

4) The length of a room is 8 ft greater than it is width. If each d imension is increased by 2 ft, the area will be increased by 60 sq. ft. Find the dimensions of the rooms.

5) The length of a rectangular lot is 4 rods greater than its wi dth, and its area is 60 square rods. Find the dimensions of the lot.

6) The length of a rectangle is 15 ft greater than its width. If each dimension is decreased by 2 ft, the area will be decreased by 106 ft 2 . Find the dimensions.

7) A rectangular piece of paper is twice as long as a square pie ce and 3 inches wider. The area of the rectangular piece is 108 in 2 . Find the dimensions of the square piece.

8) A room is one yard longer than it is wide. At 75 cper sq. yd. a covering for the oor costs S31.50. Find the dimensions of the oor.

9) The area of a rectangle is 48 ft 2 and its perimeter is 32 ft. Find its length and width.

10) The dimensions of a picture inside a frame of uniform widt h are 12 by 16 inches. If the whole area (picture and frame) is 288 in 2 , what is the width of the frame?

11) A mirror 14 inches by 15 inches has a frame of uniform width . If the area of the frame equals that of the mirror, what is the width of the fr ame.

12) A lawn is 60 ft by 80 ft. How wide a strip must be cut around it when mowing the grass to have cut half of it.

13) A grass plot 9 yards long and 6 yards wide has a path of unifo rm width around it. If the area of the path is equal to the area of the plo t, determine the width of the path.

14) A landscape architect is designing a rectangular owerb ed to be border with 28 plants that are placed 1 meter apart. He needs an inner rect angular space in the center for plants that must be 1 meter from the border of the bed and 362 that require 24 square meters for planting. What should the overall dimensions of the owerbed be?

15) A page is to have a margin of 1 inch, and is to contain 35 in 2 of painting.

How large must the page be if the length is to exceed the width b y 2 inches?

16) A picture 10 inches long by 8 inches wide has a frame whose a rea is one half the area of the picture. What are the outside dimensions of th e frame?

17) A rectangular wheat eld is 80 rods long by 60 rods wide. A s trip of uniform width is cut around the eld, so that half the grain is left sta nding in the form of a rectangular plot. How wide is the strip that is cut?

18) A picture 8 inches by 12 inches is placed in a frame of unifo rm width. If the area of the frame equals the area of the picture nd the width o f the frame.

19) A rectangular eld 225 ft by 120 ft has a ring of uniform wid th cut around the outside edge. The ring leaves 65% of the eld uncut in the c enter. What is the width of the ring?

20) One Saturday morning George goes out to cut his lot that is 100 ft by 120 ft.

He starts cutting around the outside boundary spiraling aro und towards the center. By noon he has cut 60% of the lawn. What is the width of t he ring that he has cut?

21) A frame is 15 in by 25 in and is of uniform width. The inside o f the frame leaves 75% of the total area available for the picture. What i s the width of the frame?

22) A farmer has a eld 180 ft by 240 ft. He wants to increase the area of the eld by 50% by cultivating a band of uniform width around the o utside. How wide a band should he cultivate?

23) The farmer in the previous problem has a neighber who has a eld 325 ft by 420 ft. His neighbor wants to increase the size of his eld by 2 0% by cultivating a band of uniform width around the outside of his lot. How wide a band should his neighbor cultivate?

24) A third farmer has a eld that is 500 ft by 550 ft. He wants to increase his eld by 20%. How wide a ring should he cultivate around the out side of his eld?

25) Donna has a garden that is 30 ft by 36 ft. She wants to increa se the size of the garden by 40%. How wide a ring around the outside should sh e cultivate?

26) A picture is 12 in by 25 in and is surrounded by a frame of uni form width.

The area of the frame is 30% of the area of the picture. How wide is the frame?

363