SOLUTION: I need some help, I have a test tomorrow and I don't understand some of it. It's over Quadratic Equations (solving by factoring/graphing) Here are a few problems like what I'll

Question 81910This question is from textbook Algebra 2
: I need some help, I have a test tomorrow and I don't understand some of it. It's over Quadratic Equations (solving by factoring/graphing)
Here are a few problems like what I'll have tomorrow.
* x^2-18x-40=0

* 2x^2+x-10=0
* 7x^2+6x-1=0
* (x+7)^2=49/16 This question is from textbook Algebra 2

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
x^2-18x-40=0
Think of two numbers whose product is -40 and whose sum is -18.
The numbers are -20 and +2
Rewrite the problem replacing -18x by -20x+2x
x^2-20x+2x-40=0
Factor the 1st two and the last two terms separaely to get:
x(x-20)+2(x-20)=0
Factor the left side again:
(x-20)(x+2)=0
Solve:
x=20 or x=-2
-------------
* 2x^2+x-10=0
Two numbers whose product is 2*-10=-20 and sum is 1
Numbers are -5 and +4
Rewrite as:
2x^2+5x-4x-10=0
x(2x+5)-2(2x+5)=0
(2x+5)(x-2)=0
x=-5/2 or x=2
-------------------
* 7x^2+6x-1=0
Two numbers whose product is -1*7=-7 and sum is 6
Numbers are +7 and -1
Rewrite: 7x^2+7x-x-1
7x(x+1)-(x+1)=0
(x+1)(7x-1)=0
x=-1 or x=1/7
----------------------
* (x+7)^2=49/16
This is the difference of two squares:
Rewrite as:
(x=7)^2-(49/16)=0
Factor to get:
(x-7/4)(x+7/4)=0
x=7/4 or x=-7/4
==========================
Cheers,
Stan H.
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