SOLUTION: Solve this quadratic equation using the Difference of Two Squares rule: 8x^2-2=0 Solve this quadratic equation using the "multiply to give, add to give" rule: 3x^2-5x-12=0 Th

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Solve this quadratic equation using the Difference of Two Squares rule: 8x^2-2=0 Solve this quadratic equation using the "multiply to give, add to give" rule: 3x^2-5x-12=0 Th      Log On


   

Question 225247: Solve this quadratic equation using the Difference of Two Squares rule:
8x^2-2=0
Solve this quadratic equation using the "multiply to give, add to give" rule:
3x^2-5x-12=0
Thanks
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
8x%5E2-2=0 

Divide every term by 2

8x%5E2%2F2-2%2F2=0

4x%5E2-1=0

%282x%29%5E2-1%5E2=0

%282x-1%29%282x%2B1%29=0

2x - 1 = 0      2x + 1 = 0
    2x = 1          2x = -1
     x = 1%2F2          x =-1%2F2 

-----------------------------------------------------------

red%283%29x%5E2-blue%285%29x-green%2812%29=0

Multiply red%283%29 by green%28-12%29, get -36

Think of two numbers that multiply to 
give you -36 and add to give you blue%28-5%29.

Answer: blue%28-9%29 and blue%28%22%22%2B4%29

So use those numbers blue%28-9%29 and blue%28%22%22%2B4%29

to write the middle term blue%28-5x%29 as blue%28-9x%2B4x%29

So

red%283%29x%5E2-blue%285%29x-green%2812%29=0

now becomes:

red%283%29x%5E2-blue%289x%2B4x%29-green%2812%29=0

Now out of the first two terms on the left 
factor out 3x

3x%28x-3%29%2B4x-12=0

Out of the last two terms on the left 
factor out %22%22%2B4

3x%28x-3%29%2B4%28x-3%29=0

Notice that red%28%28x-3%29%29 occurs in
both terms on the left:

3x%2Ared%28%28x-3%29%29%2B4%2Ared%28%28x-3%29%29=0

So factor out red%28%28x-3%29%29

red%28%28x-3%29%29%283x%2B4%29=0

 x - 3 = 0      3x + 4 = 0
     x = 3          3x = -4
                     x =-4%2F3
Edwin