How do you solve $x^{2} - 4 x - 7 = 0$ $x^2 - 4x - 7 = 0$ by completing the square?

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Answer 1 · 2016-10-05T14:41:42

${(x - 2)}^{2} = 11$ $(x-2)^2=11$

The process of completing the square converts the quadratic equation into a perfect square trinomial.

First move the constant, $7$ $7$ to the other side of equation by addition.

$x^{2} - 4 x = 7$ $x^2-4x=7$

Next take the coefficient of the $x$ $x$ term and divide it by 2 and then square it.

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

$4$ $4$ is amount needed to make the equation into a perfect square trinomial. Add $4$ $4$ to both sides of the equation to keep it balanced.

$x^{2} - 4 x + 4 = 7 + 4$ $x^2-4x+4=7+4$

Simplify

$x^{2} - 4 x + 4 = 11$ $x^2-4x+4=11$

Condense the equation

${(x - 2)}^{2} = 11$ $(x-2)^2=11$

Check out this tutorial on completing the square graphically.

Check out this tutorial on completing the square analytically.

How do you solve x2−4x−7=0x^2 - 4x - 7 = 0 by completing the square?