Solve for x (x^2+78x)^(1/5)=3

${(x^{2} + 78 x)}^{\frac{1}{5}} = 3$

Step 1

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

${({(x^{2} + 78 x)}^{\frac{1}{5}})}^{5} = 3^{5}$

Step 2

Simplify the exponent.

Simplify the left side.

Simplify ${({(x^{2} + 78 x)}^{\frac{1}{5}})}^{5}$ .

Multiply the exponents in ${({(x^{2} + 78 x)}^{\frac{1}{5}})}^{5}$ .

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

${(x^{2} + 78 x)}^{\frac{1}{5} \cdot 5} = 3^{5}$

Cancel the common factor of $5$ .

Cancel the common factor.

${(x^{2} + 78 x)}^{\frac{1}{5} \cdot 5} = 3^{5}$

Rewrite the expression.

${(x^{2} + 78 x)}^{1} = 3^{5}$

Simplify.

$x^{2} + 78 x = 3^{5}$

Simplify the right side.

Raise $3$ to the power of $5$ .

$x^{2} + 78 x = 243$

Step 3

Solve for $x$ .

Subtract $243$ from both sides of the equation.

$x^{2} + 78 x - 243 = 0$

Factor $x^{2} + 78 x - 243$ 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 $- 243$ and whose sum is $78$ .

$- 3, 81$

Write the factored form using these integers.

$(x - 3) (x + 81) = 0$

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

$x - 3 = 0$

$x + 81 = 0$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Set $x + 81$ equal to $0$ and solve for $x$ .

Set $x + 81$ equal to $0$ .

$x + 81 = 0$

Subtract $81$ from both sides of the equation.

$x = - 81$

The final solution is all the values that make $(x - 3) (x + 81) = 0$ true.

$x = 3, - 81$

Solve for x (x^2+78x)^(1/5)=3

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