How do you solve by completing the square $x^{2} + 2 x - 5 = 0$ $x^2+2x-5=0$ ?

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Answer 1 · 2015-06-30T20:02:37

Force a perfect square trinomial on the left side. Take the square root of both sides. Solve for $x$ $x$ , which will have two values.

Completing the square involves forcing a perfect square trinomial on the left side of the equation, then solving for $x$ $x$ .

The form for a perfect square is $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ $a^2+2ab+b^2=(a+b)^2$ .

$x^{2} + 2 x - 5 = 0$ $x^2+2x-5=0$ .

Add $5$ $5$ to both sides of the equation.

$x^{2} + 2 x = 5$ $x^2+2x=5$

Divide the coefficient of the $x$ $x$ value by $2$ $2$ , then square the result.

$\frac{2}{2} = 1;$ $2/2=1;$ $1^{2} = 1$ $1^2=1$

Add the result to both sides.

$x^{2} + 2 x + 1 = 5 + 1$ $x^2+2x+1=5+1$ =

$x^{2} + 2 x + 1 = 6$ $x^2+2x+1=6$

The left side is now a perfect square trinomial.

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

${(x + 1)}^{2} = 6$ $(x+1)^2=6$

Take the square root on both sides.

$x + 1 = \pm \sqrt{6}$ $x+1=+-sqrt6$

Subtract $1$ $1$ from both sides.

$x = \sqrt{6} - 1$ $x=sqrt6-1$

$x = - \sqrt{6} - 1$ $x=-sqrt6-1$

How do you solve by completing the square x^2+2x-5=0x2+2x−5=0x^2+2x-5=0?