1. ## finding the square

How can i find the perfect square of 2x^2-10x-7? I know how to do it is x^2 is only one, but i don't know how to solve it if it has more than 1. Solution has to be a perfect square

2. do you mean $\displaystyle (2x^2-10x-7)^2$ ???

3. sry for the late reply, but the whole equation is not squared, just the first x and you have to make it into a perfect square like 2(x-9)^2-51

4. the method is called completing the square.

$\displaystyle 2x^2 - 10x - 7$

$\displaystyle 2(x^2 - 5x) - 7$

$\displaystyle 2\left(x^2 - 5x + \frac{25}{4}\right) - 7 - \frac{25}{2}$

$\displaystyle 2\left(x - \frac{5}{2}\right)^2 - \frac{39}{2}$

5. umm im confused...where did you get 25 over 4 and 25 over 2
nvm i got it thx

P.S. i think u made a mistake . it should be 25 over 4 not 2. Ive checked many times, but not completely sure

6. Originally Posted by skeeter
the method is called completing the square.

$\displaystyle 2x^2 - 10x - 7$ This is the original equation

$\displaystyle 2(x^2 - 5x) - 7$ Here he factored a 2 out since you must have a coefficient of 1 in front of your squared term to complete the square

$\displaystyle 2\left(x^2 - 5x + \frac{25}{4}\right) - 7 - \frac{25}{2}$ the 25/4 comes from taking the coefficient of the linear term divided by 2 and then squaring that value...so (5/2)^2=(25/4)...you then add (25/4) to what you have in parenthesis and subtract the same ammount that you added times the coefficient of your quadratic equation which in this case is two so subtract (25/2)...doing this makes it so that you are really adding zero to the equation but making it easy to factor

$\displaystyle 2\left(x - \frac{5}{2}\right)^2 - \frac{39}{2}$then you just factor the quadratic
I'm sure that is tuf to follow but i will have to figure out the 'pretty print' here

7. i looked up completing the square on google and usedthis site Completing the Square. Under EXAMPLE lets try a real example. I used the way he did it and ot the answer that way hence i got 25/4 on both sides