Question

solve the cube root equation 4\sqrt3{11x - 15}+21 = 1. (1 point)
\\(x = \frac{10}{11}
\\(x = -\frac{9}{11}
\\(x = - 10
\\(x = \frac{140}{11}

Explanation:

Step1: Isolate the cube - root term

Subtract 21 from both sides of the equation $4\sqrt[3]{11x - 15}+21 = 1$.
$4\sqrt[3]{11x - 15}=1 - 21=-20$

Step2: Solve for the cube - root term

Divide both sides by 4.
$\sqrt[3]{11x - 15}=\frac{-20}{4}=-5$

Step3: Eliminate the cube - root

Cube both sides of the equation.
$(\sqrt[3]{11x - 15})^3=(-5)^3$
$11x-15=-125$

Step4: Solve for x

Add 15 to both sides: $11x=-125 + 15=-110$.
Then divide both sides by 11: $x=\frac{-110}{11}=-10$

Answer:

$x=-10$