# How do you solve x^2 + 10x + 24 = 0 using completing the square?

Jun 14, 2015

$0 = {x}^{2} + 10 x + 24 = {\left(x + 5\right)}^{2} - 1$

Hence $x = - 5 \pm \sqrt{1} = - 5 \pm 1$

#### Explanation:

${\left(x + 5\right)}^{2} = {x}^{2} + 10 x + 25$

So we have:

$0 = {x}^{2} + 10 x + 24 = {\left(x + 5\right)}^{2} - 1$

Add $1$ to both ends to get:

${\left(x + 5\right)}^{2} = 1$

So:

$x + 5 = \pm \sqrt{1} = \pm 1$

Subtract $5$ from both sides to get:

$x = - 5 \pm 1$

That is $x = - 6$ or $x = - 4$

Jun 14, 2015

Factor $y = {x}^{2} + 10 x + 24$ by completing the square

#### Explanation:

$y = {x}^{2} 10 x + \left(25 - 25\right) + 24 = 0$
$y = {\left(x + 5\right)}^{2} - 1 = 0$
${\left(x + 5\right)}^{2} = 1$ -> $x + 5 = \pm 1$

x = -5 + 1 = -4
x = -5 - 1 = -6

Jun 14, 2015

Create a perfect square trinomial on the left side of the equation, then factor it and solve for $x$. The general equation for a perfect square trinomial is ${a}^{2} + 2 a b + {b}^{2} = {\left(a + b\right)}^{2}$.

#### Explanation:

${x}^{2} + 10 x + 24 = 0$

We are going to create a perfect square trinomial on the left side of the equation, then solve for $x$. The general equation for a perfect square trinomial is ${a}^{2} + 2 a b + {b}^{2} = {\left(a + b\right)}^{2}$.

Subtract $24$ from both sides.

${x}^{2} + 10 x = - 24$

Divide the coefficient of the $x$ term by $2$, then square the result, and add it to both sides.

$\frac{10}{2} = 5$ ; 5^2=25

${x}^{2} + 10 x + 25 = - 24 + 25$ =

${x}^{2} + 10 x + 25 = 1$

We now have a perfect square trinomial on the left side, in which $a = x \mathmr{and} b = 5$. Factor the trinomial, then solve for $x$.

${\left(x + 5\right)}^{2} = 1$

Take the square root of both sides.

$x + 5 = \pm \sqrt{1}$ =

$x = \pm \sqrt{1} - 5$ =

$x = \sqrt{1} - 5 = 1 - 5 = - 4$

$x = - \sqrt{1} - 5 = - 1 - 5 = - 6$

$x = - 4$
$x = - 6$