How do you solve by completing the square for $y= (2x^2) - 4x - 7$ ?

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Answer 1 · 2015-05-25T16:41:43

I will assume that you are trying to find the values of $x$ for which the resulting $y=0$ .

$0 = y = 2x^2-4x-7 = 2(x^2-2x+1-1)-7 = 2((x-1)^2-1) -7$

$=2(x-1)^2-2-7$

$=2(x-1)^2-9$

Add $9$ to both ends to get:

$2(x-1)^2 = 9$

Divide both side by $2$ to get:

$(x-1)^2 = 9/2 = 18/4$

So $x-1 = +-sqrt(18/4) = +-(3sqrt(2))/2$

Add $1$ to both sides to get

$x = 1+-(3sqrt(2))/2$

How do you solve by completing the square for y= (2x^2) - 4x - 7y= (2x^2) - 4x - 7?