Chapter 6.7, Problem 42PPS

To determine

To Solve:

  $2{\left(x-10\right)}^{\frac{1}{3}}+4=0$

Answer to Problem 42PPS

  $x=2$

Explanation of Solution

Given Information:

Given equation:

  $2{\left(x-10\right)}^{\frac{1}{3}}+4=0$

Concept Used :

Solve Radical equations:

• Isolate the radical on one side of the equation.
• Raise each side of the equation to a power equal to the index of the radical to eliminate the radical .
• Solve the resulting polynomial equation.Check your results.
• When solving radical equations, the result may be a number that does not satisfy the original equation . Such a number is called an extraneous solution.

Calculation:

Given $2{\left(x-10\right)}^{\frac{1}{3}}+4=0$

$\begin{array}{l}2{\left(x-10\right)}^{\frac{1}{3}}+4=0\\ \Rightarrow 2{\left(x-10\right)}^{\frac{1}{3}}=0-\mathrm{4...............}[Subtract\text{ 4 from each side]}\\ \Rightarrow 2{\left(x-10\right)}^{\frac{1}{3}}=-4\\ \Rightarrow \frac{2{\left(x-10\right)}^{\frac{1}{3}}}{2}=\frac{-4}{2}\mathrm{................}[\text{Divide each side by 2]}\\ \Rightarrow {\left(x-10\right)}^{\frac{1}{3}}=-2\\ $ \Rightarrow {\left({\left(x-10\right)}^{\frac{1}{3}}\right)}^3={(-2)}^3\mathrm{...............}[\text{Raise each side to the power 3=index]}$\Rightarrow x-10=-8\\ \Rightarrow x=-8+\mathrm{10.........................}[Add\text{ 10 each side]}\\ \Rightarrow x=2\end{array}$

So, the solution is $x=2$