Question

Solve the equation with rational exponents.
left parenthesis x squared minus x minus 26 right parenthesis Superscript 3 divided by 4 Baseline minus 5 equals 3

Answer 1

Let's solve the equation:

\[
\left( x^2 - x - 26 \right)^{\frac{3}{4}} - 5 = 3
\]

First, add 5 to both sides:

\[
\left( x^2 - x - 26 \right)^{\frac{3}{4}} = 8
\]

Next, raise both sides to the power of \( \frac{4}{3} \) to eliminate the exponent:

\[
x^2 - x - 26 = 8^{\frac{4}{3}}
\]

Now, calculate \( 8^{\frac{4}{3}} \):

\[
8 = 2^3 \implies 8^{\frac{4}{3}} = (2^3)^{\frac{4}{3}} = 2^{4} = 16
\]

So we have:

\[
x^2 - x - 26 = 16
\]

Now, move 16 to the other side of the equation:

\[
x^2 - x - 16 - 26 = 0
\]
\[
x^2 - x - 42 = 0
\]

Now we can factor or use the quadratic formula. Let's use the quadratic formula where \( a = 1, b = -1, c = -42 \):

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Calculating the discriminant:

\[
b^2 - 4ac = (-1)^2 - 4(1)(-42) = 1 + 168 = 169
\]

Now, substitute back into the quadratic formula:

\[
x = \frac{1 \pm \sqrt{169}}{2}
\]
\[
x = \frac{1 \pm 13}{2}
\]

This yields two solutions:

1. \( x = \frac{14}{2} = 7 \)
2. \( x = \frac{-12}{2} = -6 \)

Thus, the solutions to the equation are:

\[
\boxed{7} \quad \text{and} \quad \boxed{-6}
\]

Let's verify these solutions in the original equation:

For \( x = 7 \):

\[
\left(7^2 - 7 - 26 \right)^{\frac{3}{4}} - 5 = (49 - 7 - 26)^{\frac{3}{4}} - 5 = (16)^{\frac{3}{4}} - 5 = 8 - 5 = 3
\]

For \( x = -6 \):

\[
\left((-6)^2 - (-6) - 26\right)^{\frac{3}{4}} - 5 = (36 + 6 - 26)^{\frac{3}{4}} - 5 = (16)^{\frac{3}{4}} - 5 = 8 - 5 = 3
\]

Both solutions satisfy the original equation, so they are confirmed correct. Final answers:

\[
\boxed{7} \quad \text{and} \quad \boxed{-6}
\]