Solve over the Interval 2(x^2-1)=10 , (0,40)

$2 (x^{2} - 1) = 10$ , $(0, 40)$

Step 1

Divide each term in $2 (x^{2} - 1) = 10$ by $2$ and simplify.

Divide each term in $2 (x^{2} - 1) = 10$ by $2$ .

$\frac{2 (x^{2} - 1)}{2} = \frac{10}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 (x^{2} - 1)}{2} = \frac{10}{2}$

Divide $x^{2} - 1$ by $1$ .

$x^{2} - 1 = \frac{10}{2}$

Simplify the right side.

Divide $10$ by $2$ .

$x^{2} - 1 = 5$

Step 2

Move all terms not containing $x$ to the right side of the equation.

Add $1$ to both sides of the equation.

$x^{2} = 5 + 1$

Add $5$ and $1$ .

$x^{2} = 6$

Step 3

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{6}$

Step 4

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = \sqrt{6}$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - \sqrt{6}$

The complete solution is the result of both the positive and negative portions of the solution.

$x = \sqrt{6}, - \sqrt{6}$

Step 5

Find the values of $n$ that produce a value within the interval $(0, 40)$ .

The interval $(0, 40)$ does not contain $- \sqrt{6}$ . It is not part of the final solution.

$- \sqrt{6}$ is not on the interval

The interval $(0, 40)$ contains $\sqrt{6}$ .

$x = \sqrt{6}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = \sqrt{6}$

Decimal Form:

$x = 2.44948974 \dots$