SOLUTION: Hi! I have a homework that says: solve the following quadratic equations by factoring. a) x^2+8x+7=0 I tried to solve it and I got this answer (x+1) (x+7)=0. b) 2y^2 + 12

Algebra ->  Test -> SOLUTION: Hi! I have a homework that says: solve the following quadratic equations by factoring. a) x^2+8x+7=0 I tried to solve it and I got this answer (x+1) (x+7)=0. b) 2y^2 + 12      Log On


   

Question 596729: Hi!
I have a homework that says:
solve the following quadratic equations by factoring.
a) x^2+8x+7=0 I tried to solve it and I got this answer (x+1) (x+7)=0.
b) 2y^2 + 12y +10=0 I got: 2(y^2+6y+5)=0
c) x^2=81 I'm not sure how o solve this one.
However the =0 confuses me do I have to use the quadratic formula to solve them?
Thank you.
Answer by KMST(5399) About Me  (Show Source):
You can put this solution on YOUR website!
I would not use the quadratic formula unless I really had to, and the instructions say to use factoring.

a) x%5E2%2B8x%2B7=0 You factored the quadratic polynomial to get %28x%2B1%29+%28x%2B7%29=0.
For the product to be zero, one factor must be zero.
x%2B1=0 gives you one solution: highlight%28x=-1%29
x%2B7=0 gives you the other solution: highlight%28x=-7%29

b) 2y%5E2+%2B+12y+%2B10=0 You factored the quadratic polynomial to get 2%28y%5E2%2B6y%2B5%29=0
Factoring a little further, you get 2%28y%2B1%29%28y%2B5%29=0
One of those factors must be zero.
y%2B1=0 gives you one solution: highlight%28y=-1%29
y%2B5=0 gives you the other solution: highlight%28y=-5%29

c) x%5E2=81 is easy, except for the question of using factoring.
x%5E2=81 means that x is a number that squared equals 81.
You know one such number: 9%5E2=81 and highlight%28x=9%29 is one solution.
There is also highlight%28x=-9%29 because %28-9%29%5E2=81 too.
Maybe you were expected to transform it by subtracting 81 from both sides to get
x%5E2-81=0 and then factor to get %28x-9%29%28x%2B9%29=0.