Solve for x (x^2+14x)^(1/5)=2

${(x^{2} + 14 x)}^{\frac{1}{5}} = 2$

Raise each side of the equation to the power of $5$ to eliminate the fractional exponent on the left side.

${({(x^{2} + 14 x)}^{\frac{1}{5}})}^{5} = 2^{5}$

Simplify the exponent.

Simplify the left side.

Simplify ${({(x^{2} + 14 x)}^{\frac{1}{5}})}^{5}$ .

Multiply the exponents in ${({(x^{2} + 14 x)}^{\frac{1}{5}})}^{5}$ .

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

${(x^{2} + 14 x)}^{\frac{1}{5} \cdot 5} = 2^{5}$

Cancel the common factor of $5$ .

Cancel the common factor.

${(x^{2} + 14 x)}^{\frac{1}{5} \cdot 5} = 2^{5}$

Rewrite the expression.

${(x^{2} + 14 x)}^{1} = 2^{5}$

Simplify.

$x^{2} + 14 x = 2^{5}$

Simplify the right side.

Raise $2$ to the power of $5$ .

$x^{2} + 14 x = 32$

Solve for $x$ .

Subtract $32$ from both sides of the equation.

$x^{2} + 14 x - 32 = 0$

Factor $x^{2} + 14 x - 32$ 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 $- 32$ and whose sum is $14$ .

$- 2, 16$

Write the factored form using these integers.

$(x - 2) (x + 16) = 0$

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

$x - 2 = 0$

$x + 16 = 0$

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Set $x + 16$ equal to $0$ and solve for $x$ .

Set $x + 16$ equal to $0$ .

$x + 16 = 0$

Subtract $16$ from both sides of the equation.

$x = - 16$

The final solution is all the values that make $(x - 2) (x + 16) = 0$ true.

$x = 2, - 16$

Solve for x (x^2+14x)^(1/5)=2

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