Solve for x square root of x^2-6x-6-3=4

$\sqrt{x^{2} - 6 x - 6} - 3 = 4$

Step 1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x^{2} - 6 x - 6}$ as ${(x^{2} - 6 x - 6)}^{\frac{1}{2}}$ .

${(x^{2} - 6 x - 6)}^{\frac{1}{2}} - 3 = 4$

Step 2

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

Add $3$ to both sides of the equation.

${(x^{2} - 6 x - 6)}^{\frac{1}{2}} = 4 + 3$

Add $4$ and $3$ .

${(x^{2} - 6 x - 6)}^{\frac{1}{2}} = 7$

Step 3

Raise each side of the equation to the power of $2$ to eliminate the fractional exponent on the left side.

${({(x^{2} - 6 x - 6)}^{\frac{1}{2}})}^{2} = 7^{2}$

Step 4

Simplify the exponent.

Simplify the left side.

Simplify ${({(x^{2} - 6 x - 6)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(x^{2} - 6 x - 6)}^{\frac{1}{2}})}^{2}$ .

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

${(x^{2} - 6 x - 6)}^{\frac{1}{2} \cdot 2} = 7^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(x^{2} - 6 x - 6)}^{\frac{1}{2} \cdot 2} = 7^{2}$

Rewrite the expression.

${(x^{2} - 6 x - 6)}^{1} = 7^{2}$

Simplify.

$x^{2} - 6 x - 6 = 7^{2}$

Simplify the right side.

Raise $7$ to the power of $2$ .

$x^{2} - 6 x - 6 = 49$

Step 5

Solve for $x$ .

Subtract $49$ from both sides of the equation.

$x^{2} - 6 x - 6 - 49 = 0$

Subtract $49$ from $- 6$ .

$x^{2} - 6 x - 55 = 0$

Factor $x^{2} - 6 x - 55$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 55$ and whose sum is $- 6$ .

$- 11, 5$

Write the factored form using these integers.

$(x - 11) (x + 5) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 11 = 0$

$x + 5 = 0$

Set $x - 11$ equal to $0$ and solve for $x$ .

Set $x - 11$ equal to $0$ .

$x - 11 = 0$

Add $11$ to both sides of the equation.

$x = 11$

Set $x + 5$ equal to $0$ and solve for $x$ .

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

The final solution is all the values that make $(x - 11) (x + 5) = 0$ true.

$x = 11, - 5$

Solve for x square root of x^2-6x-6-3=4

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