Q. 8.118 - Intermediate Algebra

Question

Solve: $\sqrt[3]{6x-10}+1=-3$.

Step-by-Step Solution

Answer

The solution is, $x=-9$.

Step 1 Given equation is,

$\sqrt[3]{6x-10}+1=-3$.

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

Step 2 To isolate the radical 1 on both sides.

$$\begin{gathered} \sqrt[3]{6x-10}+1-1=-3-1 \\ \sqrt[3]{6x-10}=-4 \end{gathered}$$

Now take cube on both sides.

$$\begin{gathered} 6x-10={\left(-4\right)}^3 \\ 6x-10=-64 \end{gathered}$$

Now simplify the obtained equation.

$$\begin{gathered} 6x=-64+10 \\ 6x=-54 \\ x=\frac{-54}{6} \\ x=-9 \end{gathered}$$

Step 3 Let us check the answer.

Substitute $x=-9$ into the equation $\sqrt[3]{6x-10}+1=-3$.

$$\begin{gathered} \sqrt[3]{6\left(-9\right)-10}+1=-3 \\ \sqrt[3]{-54-10}+1=-3 \\ \sqrt[3]{-64}+1=-3 \\ \sqrt[3]{{\left(-4\right)}^3}+1=-3 \\ -4+1=-3 \\ -3=-3 \end{gathered}$$

which is true.