Solve over the Interval x(x+4)=24 , (-2,9)

$x (x + 4) = 24$ , $(- 2, 9)$

Step 1

Simplify $x (x + 4)$ .

Apply the distributive property.

$x \cdot x + x \cdot 4 = 24$

Simplify the expression.

Multiply $x$ by $x$ .

$x^{2} + x \cdot 4 = 24$

Move $4$ to the left of $x$ .

$x^{2} + 4 x = 24$

Step 2

Subtract $24$ from both sides of the equation.

$x^{2} + 4 x - 24 = 0$

Step 3

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 4

Substitute the values $a = 1$ , $b = 4$ , and $c = - 24$ into the quadratic formula and solve for $x$ .

$\frac{- 4 \pm \sqrt{4^{2} - 4 \cdot (1 \cdot - 24)}}{2 \cdot 1}$

Step 5

Simplify.

Simplify the numerator.

Raise $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot - 24}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 24$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot - 24}}{2 \cdot 1}$

Multiply $- 4$ by $- 24$ .

$x = \frac{- 4 \pm \sqrt{16 + 96}}{2 \cdot 1}$

Add $16$ and $96$ .

$x = \frac{- 4 \pm \sqrt{112}}{2 \cdot 1}$

Rewrite $112$ as $4^{2} \cdot 7$ .

Factor $16$ out of $112$ .

$x = \frac{- 4 \pm \sqrt{16 (7)}}{2 \cdot 1}$

Rewrite $16$ as $4^{2}$ .

$x = \frac{- 4 \pm \sqrt{4^{2} \cdot 7}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 4 \pm 4 \sqrt{7}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 4 \pm 4 \sqrt{7}}{2}$

Simplify $\frac{- 4 \pm 4 \sqrt{7}}{2}$ .

$x = - 2 \pm 2 \sqrt{7}$

Step 6

Simplify the expression to solve for the $+$ portion of the $\pm$ .

Simplify the numerator.

Raise $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot - 24}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 24$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot - 24}}{2 \cdot 1}$

Multiply $- 4$ by $- 24$ .

$x = \frac{- 4 \pm \sqrt{16 + 96}}{2 \cdot 1}$

Add $16$ and $96$ .

$x = \frac{- 4 \pm \sqrt{112}}{2 \cdot 1}$

Rewrite $112$ as $4^{2} \cdot 7$ .

Factor $16$ out of $112$ .

$x = \frac{- 4 \pm \sqrt{16 (7)}}{2 \cdot 1}$

Rewrite $16$ as $4^{2}$ .

$x = \frac{- 4 \pm \sqrt{4^{2} \cdot 7}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 4 \pm 4 \sqrt{7}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 4 \pm 4 \sqrt{7}}{2}$

Simplify $\frac{- 4 \pm 4 \sqrt{7}}{2}$ .

$x = - 2 \pm 2 \sqrt{7}$

Change the $\pm$ to $+$ .

$x = - 2 + 2 \sqrt{7}$

Step 7

Simplify the expression to solve for the $-$ portion of the $\pm$ .

Simplify the numerator.

Raise $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot - 24}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 24$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot - 24}}{2 \cdot 1}$

Multiply $- 4$ by $- 24$ .

$x = \frac{- 4 \pm \sqrt{16 + 96}}{2 \cdot 1}$

Add $16$ and $96$ .

$x = \frac{- 4 \pm \sqrt{112}}{2 \cdot 1}$

Rewrite $112$ as $4^{2} \cdot 7$ .

Factor $16$ out of $112$ .

$x = \frac{- 4 \pm \sqrt{16 (7)}}{2 \cdot 1}$

Rewrite $16$ as $4^{2}$ .

$x = \frac{- 4 \pm \sqrt{4^{2} \cdot 7}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 4 \pm 4 \sqrt{7}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 4 \pm 4 \sqrt{7}}{2}$

Simplify $\frac{- 4 \pm 4 \sqrt{7}}{2}$ .

$x = - 2 \pm 2 \sqrt{7}$

Change the $\pm$ to $-$ .

$x = - 2 - 2 \sqrt{7}$

Step 8

The final answer is the combination of both solutions.

$x = - 2 + 2 \sqrt{7}, - 2 - 2 \sqrt{7}$

Step 9

Find the values of $n$ that produce a value within the interval $(- 2, 9)$ .

The interval $(- 2, 9)$ does not contain $- 2 - 2 \sqrt{7}$ . It is not part of the final solution.

$- 2 - 2 \sqrt{7}$ is not on the interval

The interval $(- 2, 9)$ contains $- 2 + 2 \sqrt{7}$ .

$x = - 2 + 2 \sqrt{7}$

Step 10

The result can be shown in multiple forms.

Exact Form:

$x = - 2 + 2 \sqrt{7}$

Decimal Form:

$x = 3.29150262 \dots$

Solve over the Interval x(x+4)=24 , (-2,9)

Related Questions

Support LCMGCF.com by disabling your adblocker.