Question

what is the solution to the equation \\(sqrt3{x + 4} + sqrt3{2x + 8} = 0\\)?
options:
\\(x = -4\\)
\\(x = -12\\)
\\(x = 4\\)
\\(x = 12\\)

Explanation:

Step1: Analyze the equation

We have the equation \(\sqrt[3]{x + 4}+\sqrt[3]{2x+8}=0\). Let's rewrite \(\sqrt[3]{2x + 8}\) as \(\sqrt[3]{2(x + 4)}\) or we can use the property that if \(a + b=0\), then \(a=-b\). So, \(\sqrt[3]{x + 4}=-\sqrt[3]{2x + 8}\).

Step2: Cube both sides

Cubing both sides of the equation \(\sqrt[3]{x + 4}=-\sqrt[3]{2x + 8}\), we know that \((\sqrt[3]{a})^3=a\) and \((-b)^3=-b^3\). So, \((\sqrt[3]{x + 4})^3=(-\sqrt[3]{2x + 8})^3\) which simplifies to \(x + 4=-(2x + 8)\).

Step3: Solve the linear equation

Expand the right - hand side: \(x + 4=-2x-8\).
Add \(2x\) to both sides: \(x+2x + 4=-2x + 2x-8\), so \(3x+4=-8\).
Subtract 4 from both sides: \(3x+4 - 4=-8 - 4\), so \(3x=-12\).
Divide both sides by 3: \(x=\frac{-12}{3}=-4\).

Step4: Verify the solution

Substitute \(x = - 4\) into the original equation:
Left - hand side: \(\sqrt[3]{-4 + 4}+\sqrt[3]{2\times(-4)+8}=\sqrt[3]{0}+\sqrt[3]{-8 + 8}=0 + 0=0\).
The right - hand side is 0. So \(x=-4\) is a valid solution.

Answer:

\(x=-4\) (corresponding to the option with \(x = - 4\))