Solve for x square root of x^2-4=2 square root of 3

$\sqrt{x^{2} - 4} = 2 \sqrt{3}$

Step 1

To remove the radical on the left side of the equation, square both sides of the equation.

${\sqrt{x^{2} - 4}}^{2} = {(2 \sqrt{3})}^{2}$

Step 2

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x^{2} - 4}$ as ${(x^{2} - 4)}^{\frac{1}{2}}$ .

${({(x^{2} - 4)}^{\frac{1}{2}})}^{2} = {(2 \sqrt{3})}^{2}$

Simplify the left side.

Simplify ${({(x^{2} - 4)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(x^{2} - 4)}^{\frac{1}{2}})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(x^{2} - 4)}^{\frac{1}{2} \cdot 2} = {(2 \sqrt{3})}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(x^{2} - 4)}^{\frac{1}{2} \cdot 2} = {(2 \sqrt{3})}^{2}$

Rewrite the expression.

${(x^{2} - 4)}^{1} = {(2 \sqrt{3})}^{2}$

Simplify.

$x^{2} - 4 = {(2 \sqrt{3})}^{2}$

Simplify the right side.

Simplify ${(2 \sqrt{3})}^{2}$ .

Simplify the expression.

Apply the product rule to $2 \sqrt{3}$ .

$x^{2} - 4 = 2^{2} {\sqrt{3}}^{2}$

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

$x^{2} - 4 = 4 {\sqrt{3}}^{2}$

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$x^{2} - 4 = 4 {(3^{\frac{1}{2}})}^{2}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$x^{2} - 4 = 4 \cdot 3^{\frac{1}{2} \cdot 2}$

Combine $\frac{1}{2}$ and $2$ .

$x^{2} - 4 = 4 \cdot 3^{\frac{2}{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$x^{2} - 4 = 4 \cdot 3^{\frac{2}{2}}$

Rewrite the expression.

$x^{2} - 4 = 4 \cdot 3^{1}$

Evaluate the exponent.

$x^{2} - 4 = 4 \cdot 3$

Multiply $4$ by $3$ .

$x^{2} - 4 = 12$

Step 3

Solve for $x$ .

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

Add $4$ to both sides of the equation.

$x^{2} = 12 + 4$

Add $12$ and $4$ .

$x^{2} = 16$

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

$x = \pm \sqrt{16}$

Simplify $\pm \sqrt{16}$ .

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

$x = \pm \sqrt{4^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 4$

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.

$x = 4$

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

$x = - 4$

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

$x = 4, - 4$

Solve for x square root of x^2-4=2 square root of 3

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