Solve for x (x^2+24x)^(1/4)=3

${(x^{2} + 24 x)}^{\frac{1}{4}} = 3$

Raise each side of the equation to the power of $4$ to eliminate the fractional exponent on the left side.

${({(x^{2} + 24 x)}^{\frac{1}{4}})}^{4} = 3^{4}$

Simplify the exponent.

Simplify the left side.

Simplify ${({(x^{2} + 24 x)}^{\frac{1}{4}})}^{4}$ .

Multiply the exponents in ${({(x^{2} + 24 x)}^{\frac{1}{4}})}^{4}$ .

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

${(x^{2} + 24 x)}^{\frac{1}{4} \cdot 4} = 3^{4}$

Cancel the common factor of $4$ .

Cancel the common factor.

${(x^{2} + 24 x)}^{\frac{1}{4} \cdot 4} = 3^{4}$

Rewrite the expression.

${(x^{2} + 24 x)}^{1} = 3^{4}$

Simplify.

$x^{2} + 24 x = 3^{4}$

Simplify the right side.

Raise $3$ to the power of $4$ .

$x^{2} + 24 x = 81$

Solve for $x$ .

Subtract $81$ from both sides of the equation.

$x^{2} + 24 x - 81 = 0$

Factor $x^{2} + 24 x - 81$ 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 $- 81$ and whose sum is $24$ .

$- 3, 27$

Write the factored form using these integers.

$(x - 3) (x + 27) = 0$

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

$x - 3 = 0$

$x + 27 = 0$

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 $x + 27$ equal to $0$ and solve for $x$ .

Set $x + 27$ equal to $0$ .

$x + 27 = 0$

Subtract $27$ from both sides of the equation.

$x = - 27$

The final solution is all the values that make $(x - 3) (x + 27) = 0$ true.

$x = 3, - 27$

Solve for x (x^2+24x)^(1/4)=3

Related Questions

Support LCMGCF.com by disabling your adblocker.