Week 4: Student Discussion and Response

YesterdayAug 11 at 10:11am

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The two equations I needed to solve were x2−2x−24=0 (which needed to be solved using factoring) and (x+5)2=4 (which required the use of the quadratic formula). The following are solutions to both of these dastardly situations:

x2−2x−24=0 I began the solution by factoring out the starting equation, finding two numbers

(x−6)(x+4)=0 whose product was 24 but also -2 when added together. We split the two factors

x - 6 = 0 x + 4 = 0 and solve for each, giving us x=6 and x=-4. Thus, the solution is x = {6, -4}

x = 6 and x = -4 If we were to solve this by completing the square, in the beginning we would

x={6,−4} have divided each term by 1 (for x) but it wouldn't have been efficient..... I think.

(x+5)2=4 For this one, we need to solve using the quadratic formula. So we begin by turning

(x+5)(x+5)=4 (x+5)^2 into (x+5)(x+5) and factoring on the left hand side using FOIL which leaves

x2+5x+5x+10=4 a bunch of new terms to combine and then subtract 4 from the right side to

x2+10x+6=0 get 0, which finally allows us to convert the whole situation into the quadratic

x=10±(10)2−4(1)(6)2(1) formula. Our discriminant is (10)^2 - 4(1)(6) which is under the square root. x=−10±762 After multiplying then subtracting, we get -10 plus/minus the square root of 76. We x=−5±192 then split 76 into the square root of 4 times the square root of 19. We further solve

x=−5+192 the equation in -10 plus/minus 2 times the square root of 19, then we divide each

x=−5−192 term by 2, leaving x=-5 plus/minus the square root of 19 divided by 2. So, our

x=2.82 x=-7.18 solution becomes both x equals -5 plus the square root of 19 over 2 and x equals -5

minus the square root of 19 over 2, leaving x=2.82 and x=-7.18.