Q. 329 - Intermediate Algebra

Question

In the following exercises, solve.
$\sqrt[3]{2x^2+9x-18}=\sqrt[3]{x^2+3x-2}$

Step-by-Step Solution

Answer

The solutions of the given equation are $x=-8$ and $x=2$.

Step 1. Given information.

The given equation is:

$\sqrt[3]{2x^2+9x-18}=\sqrt[3]{x^2+3x-2}$

Step 2. Simplify the equation.

Take cube on both sides.

$\begin{array}{rcl}(\sqrt[3]{2x^2+9x-18}{)}^3& =& (\sqrt[3]{x^2+3x-2}{)}^3\\ 2x^2+9x-18& =& x^2+3x-2\\ 2x^2+9x-18-x^2-3x+2& =& 0\\ x^2+6x-16& =& 0\end{array}$

Step 3. Solve the equation.

Splitting the middle term, the above equation can be written as:

$\begin{array}{rcl}{x}^2+8x-2x-16& =& 0\\ x(x+8)-2(x+8)& =& 0\\ (x+8)(x-2)& =& 0\\ x& =& -8,2\end{array}$

Step 4. Conclusion.

The solutions of the given equation are $x=-8$ and $x=2$.