Q. 8.117 - Intermediate Algebra

Question

Solve: $\sqrt[3]{4x-3}+8=5$.

Step-by-Step Solution

Answer

The solution is, $x=-6$.

Step 1 Given equation is,

$\sqrt[3]{4x-3}+8=5$.

The objective is, we need to find the solution for the given equation.

Step 2   To isolate the radical, subtract 8 on both sides.

$$\begin{gathered} \sqrt[3]{4x-3}+8-8=5-8 \\ \sqrt[3]{4x-3}=-3 \end{gathered}$$

Now take cube on both sides.

$4x-3={\left(-3\right)}^3$

Now simplify the above equation.

$$\begin{gathered} 4x-3=-27 \\ 4x=-27+3 \\ 4x=-24 \\ x=\frac{-24}{4} \\ x=-6 \end{gathered}$$

Step 3 Let us check the answer.

Substitute $x=-6$ into the equation. $$\begin{gathered} \sqrt[3]{4\left(-6\right)-3}+8=5 \\ \sqrt[3]{-24-3}+8=5 \\ \sqrt[3]{-27}+8=5 \\ \sqrt[3]{{\left(-3\right)}^3}+8=5 \\ -3+8=5 \\ 5=5 \end{gathered}$$

which is true.