what is the solution to the equation $sqrt3{x + 4}+sqrt3{2x + 8}=0$?\n$x=-12$\n$x = - 4$\n$x = 4$\n$x = 12$
what is the solution to the equation $sqrt3{x + 4}+sqrt3{2x + 8}=0$?\n$x=-12$\n$x = - 4$\n$x = 4$\n$x = 12$
Answer
Explanation:
Step1: Isolate one cube - root term
Subtract $\sqrt[3]{2x + 8}$ from both sides of the equation $\sqrt[3]{x + 4}+\sqrt[3]{2x + 8}=0$. We get $\sqrt[3]{x + 4}=-\sqrt[3]{2x + 8}$.
Step2: Cube both sides
Cubing both sides of the equation $\sqrt[3]{x + 4}=-\sqrt[3]{2x + 8}$, we have $(\sqrt[3]{x + 4})^3=(-\sqrt[3]{2x + 8})^3$. Since $(\sqrt[3]{a})^3=a$ and $(-\sqrt[3]{b})^3=-b$, the equation becomes $x + 4=-(2x + 8)$.
Step3: Expand and solve for x
Expand the right - hand side: $x + 4=-2x-8$. Add $2x$ to both sides: $x+2x + 4=-2x+2x-8$, which simplifies to $3x + 4=-8$. Subtract 4 from both sides: $3x+4 - 4=-8 - 4$, so $3x=-12$. Divide both sides by 3: $\frac{3x}{3}=\frac{-12}{3}$, and $x=-4$.
Answer:
$x = - 4$ (B. $x=-4$)