Week 4: Student Discussion and Response
Patrick WarthenThursdayAug 10 at 3:15pm
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For this assignment, I am to use factoring to solve x^2+6x+5=0 and the quadratic formula to solve 2x^2 –x=6
To factor the first problem I must separate the expressions into groups.
I do this by dropping the zero at the end, as it doesn’t have much use at the moment.
(x^2+x)+(5x+5)
Then I factor out x from x^2+x which gives me x(x+1).
Next is to factor our 5 from 5x+5 which gives me 5(x+1).
The end result being x(x+1)+5(x+1).
With this we can factor out the common term of (x+1), so we end up with (x+1)(x+5).
Add the zero back and the answer is (x+1)(x+5)=0
Zero Factor Principle time. x+1=0, so x must be -1. x+5=0, so x must be -5.
The solutions are therefore x=-1, x=-5.
The next is to be solved via the quadratic formula, which is more difficult than completing the square but it must be done.
Starting off with 2x^2 –x=6 I apply the formula as ax^2+bx+c=0 and end up with a=2, b=-1, and c=-6.
I applied the rule –(-a)=a, multiplies the numbers, found the square root of 49 along the way.
I got the results by figuring –(-1)+√((-1)^2-4*2(-6)) over 2*2. Using the tried and true PEMDAS I got 2 and –(3/2) which is negative three halves.
Oh yeah, the discriminant was √((-1)^2-4*2(-6)).
Hope that makes any sense at all, but x=2, x=-(3/2).