Solve for x (x^2-x-4)^(3/4)-2=6

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

Step 1

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

Add $2$ to both sides of the equation.

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

Add $6$ and $2$ .

${(x^{2} - x - 4)}^{\frac{3}{4}} = 8$

Step 2

Raise each side of the equation to the power of $\frac{4}{3}$ to eliminate the fractional exponent on the left side.

${({(x^{2} - x - 4)}^{\frac{3}{4}})}^{\frac{4}{3}} = 8^{\frac{4}{3}}$

Step 3

Simplify the exponent.

Simplify the left side.

Simplify ${({(x^{2} - x - 4)}^{\frac{3}{4}})}^{\frac{4}{3}}$ .

Multiply the exponents in ${({(x^{2} - x - 4)}^{\frac{3}{4}})}^{\frac{4}{3}}$ .

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

${(x^{2} - x - 4)}^{\frac{3}{4} \cdot \frac{4}{3}} = 8^{\frac{4}{3}}$

Cancel the common factor of $3$ .

Cancel the common factor.

${(x^{2} - x - 4)}^{\frac{3}{4} \cdot \frac{4}{3}} = 8^{\frac{4}{3}}$

Rewrite the expression.

${(x^{2} - x - 4)}^{\frac{1}{4} \cdot 4} = 8^{\frac{4}{3}}$

Cancel the common factor of $4$ .

Cancel the common factor.

${(x^{2} - x - 4)}^{\frac{1}{4} \cdot 4} = 8^{\frac{4}{3}}$

Rewrite the expression.

${(x^{2} - x - 4)}^{1} = 8^{\frac{4}{3}}$

Simplify.

$x^{2} - x - 4 = 8^{\frac{4}{3}}$

Simplify the right side.

Simplify $8^{\frac{4}{3}}$ .

Simplify the expression.

Rewrite $8$ as $2^{3}$ .

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

$x^{2} - x - 4 = 2^{4}$

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

$x^{2} - x - 4 = 16$

Step 4

Solve for $x$ .

Subtract $16$ from both sides of the equation.

$x^{2} - x - 4 - 16 = 0$

Subtract $16$ from $- 4$ .

$x^{2} - x - 20 = 0$

Factor $x^{2} - x - 20$ 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 $- 20$ and whose sum is $- 1$ .

$- 5, 4$

Write the factored form using these integers.

$(x - 5) (x + 4) = 0$

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

$x - 5 = 0$

$x + 4 = 0$

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

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

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Set $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

The final solution is all the values that make $(x - 5) (x + 4) = 0$ true.

$x = 5, - 4$

Solve for x (x^2-x-4)^(3/4)-2=6

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