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

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Answer 1 · 2015-05-25T16:41:43

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

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

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

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

Add $9$ $9$ to both ends to get:

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

Divide both side by $2$ $2$ to get:

${(x - 1)}^{2} = \frac{9}{2} = \frac{18}{4}$ $(x-1)^2 = 9/2 = 18/4$

So $x - 1 = \pm \sqrt{\frac{18}{4}} = \pm \frac{3 \sqrt{2}}{2}$ $x-1 = +-sqrt(18/4) = +-(3sqrt(2))/2$

Add $1$ $1$ to both sides to get

$x = 1 \pm \frac{3 \sqrt{2}}{2}$ $x = 1+-(3sqrt(2))/2$

How do you solve by completing the square for y=(2x2)−4x−7y= (2x^2) - 4x - 7?