How do you solve by completing the square: $x^{2} - 4 x - 11 = 0$ $x^2- 4x-11=0$ ?

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Answer 1 · 2015-04-05T07:50:29

First, we Transpose the Constant to one side of the equation.
Transposing $- 11$ $-11$ to the other side we get:
$x^{2} - 4 x = 11$ $x^2-4x = 11$
Application of ${(a - b)}^{2} = a^{2} - 2 a b + b^{2}$ $(a-b)^2 = a^2 - 2ab + b^2$
We look at the Co-efficient of $x$ $x$ . It's $- 4$ $-4$
We take half of this number (including the sign), giving us $–2$
We square this value to get $(-2)^2 = 4$ . We add this number to BOTH sides of the Equation.
$x^2-4x+4 = 11+4$
$x^2-4x+4 = 15$
The Left Hand side $x^2-4x+4$ is in the form $a^2 - 2ab + b^2$
where $a$ is $x$ , and $b$ is $2$
The equation can be written as
$(x-2)^2 = 15$

So $(x-2)$ can take either $sqrt15$ or $-sqrt15$ as a value. That's because squaring either will give us 15.

$x-2 = sqrt15$ (or) $x-2 = -sqrt15$
$x = 2+sqrt15$ (or) $x = 2-sqrt15$

Solution : $x = 2+sqrt15,2-sqrt15$
Verify your answer by substituting these values in the Original Equation $x^2- 4x - 11 = 0$
You will see that the Solution is correct.

How do you solve by completing the square: x^2- 4x-11=0x2−4x−11=0x^2- 4x-11=0?