How do you solve $x^{2} - 4 x = - 3$ $x^2 - 4x = -3$ by completing the square?

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Answer 1 · 2017-10-11T00:20:00

$x = 3$ $x=3$ OR $x = 1$ $x=1$

We want to add a number on both sides of the equation so that the left side becomes a perfect square:

To do that, we look at $b$ $b$ (the number next to $x$ $x$ ) and see that it is $- 4$ $-4$

If you add ${(\frac{b}{2})}^{2}$ $(b/2)^2$ to both sides, the equation on the left will be a perfect square or:

${(- \frac{4}{2})}^{2} = {(- 2)}^{2} = 4$ $(-4/2)^2=(-2)^2=4$

So:

$x^{2} - 4 x + 4 = - 3 + 4$ $x^2-4xcolor(red)(+4)=-3color(red)(+4)$

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

$\sqrt{{(x - 2)}^{2}} = \sqrt{1}$ $sqrt((x-2)^2)=sqrt(1)$

$x - 2 = 1$ $x-2=1$ OR $x - 2 = - 1$ $x-2=-1$

$x = 3$ $x=3$ OR $x = 1$ $x=1$

How do you solve x^2 - 4x = -3x2−4x=−3x^2 - 4x = -3 by completing the square?