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

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Answer 1 · 2015-08-01T13:13:46

$x_{1, 2} = - 1 \pm \sqrt{6}$ $x_(1,2) = -1 +- sqrt(6)$

First, get your quadratic equation to the form

$x^{2} + \frac{b}{a} x = - \frac{c}{a}$ $color(blue)(x^2 + b/ax = -c/a)$

To do that, add $5$ $5$ to both sides of the equation

$x^2 + 2x - color(red)(cancel(color(black)(5))) + color(red)(cancel(color(black)(5))) = 5$

$x^2 + 2x = 5$

Next, divide the coefficient of the $x$ -term by 2, square the result. then add it to both sides of the equation.

$2/2 = 1$ , then $1^2 = 1$

This will get you

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

Notice that you can rewrite the left side of the equation as the square of a binomial

$x^2 + 2x + 1 = x^2 + 2 * (1) * x + 1^2 = (x+1)^2$

You will now have

$(x + 1)^2 = 6$

Take the square root of both sides

$sqrt((x+1)^2) = sqrt(6)$

$x + 1 = +- sqrt(6) => x_(1,2) = -1 +- sqrt(6) = {(x_1 = color(green)(-1 - sqrt(6))), (x_2 = color(green)(-1 + sqrt(6))) :}$

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