Answer :
To determine which statement is true about the solution of the equation [tex]\(\sqrt[3]{x^2 - 12} = \sqrt[3]{4x}\)[/tex], let's go through the steps to find and verify the solutions.
Step-by-Step Solution:
1. Set Up the Equation:
[tex]\[\sqrt[3]{x^2 - 12} = \sqrt[3]{4x}\][/tex]
2. Eliminate the Cube Roots:
To eliminate the cube roots, we can cube both sides of the equation:
[tex]\[(\sqrt[3]{x^2 - 12})^3 = (\sqrt[3]{4x})^3\][/tex]
This simplifies to:
[tex]\[x^2 - 12 = 4x\][/tex]
3. Rearrange the Equation:
Let's arrange the equation in standard quadratic form by moving all terms to one side:
[tex]\[x^2 - 4x - 12 = 0\][/tex]
4. Solve the Quadratic Equation:
We can solve the quadratic equation using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = -4\)[/tex], and [tex]\(c = -12\)[/tex]:
[tex]\[x = \frac{4 \pm \sqrt{16 + 48}}{2}\][/tex]
[tex]\[x = \frac{4 \pm \sqrt{64}}{2}\][/tex]
[tex]\[x = \frac{4 \pm 8}{2}\][/tex]
This gives us two solutions:
[tex]\[x = \frac{4 + 8}{2} = 6 \][/tex]
[tex]\[x = \frac{4 - 8}{2} = -2\][/tex]
5. Verify the Solutions:
We need to check if these solutions satisfy the original equation.
- For [tex]\(x = 6\)[/tex]:
Substitute [tex]\(x = 6\)[/tex] into the original equation:
[tex]\[\sqrt[3]{6^2 - 12} = \sqrt[3]{4 \cdot 6}\][/tex]
[tex]\[\sqrt[3]{36 - 12} = \sqrt[3]{24}\][/tex]
[tex]\[\sqrt[3]{24} = \sqrt[3]{24}\][/tex]
This is true.
- For [tex]\(x = -2\)[/tex]:
Substitute [tex]\(x = -2\)[/tex] into the original equation:
[tex]\[\sqrt[3]{(-2)^2 - 12} = \sqrt[3]{4 \cdot (-2)}\][/tex]
[tex]\[\sqrt[3]{4 - 12} = \sqrt[3]{-8}\][/tex]
[tex]\[\sqrt[3]{-8} = \sqrt[3]{-8}\][/tex]
This is true.
Since both [tex]\(x = -2\)[/tex] and [tex]\(x = 6\)[/tex] satisfy the original equation, we conclude that:
[tex]\[ \boxed{\text{Both } x = -2 \text{ and } x = 6 \text{ are true solutions.}} \][/tex]
Step-by-Step Solution:
1. Set Up the Equation:
[tex]\[\sqrt[3]{x^2 - 12} = \sqrt[3]{4x}\][/tex]
2. Eliminate the Cube Roots:
To eliminate the cube roots, we can cube both sides of the equation:
[tex]\[(\sqrt[3]{x^2 - 12})^3 = (\sqrt[3]{4x})^3\][/tex]
This simplifies to:
[tex]\[x^2 - 12 = 4x\][/tex]
3. Rearrange the Equation:
Let's arrange the equation in standard quadratic form by moving all terms to one side:
[tex]\[x^2 - 4x - 12 = 0\][/tex]
4. Solve the Quadratic Equation:
We can solve the quadratic equation using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = -4\)[/tex], and [tex]\(c = -12\)[/tex]:
[tex]\[x = \frac{4 \pm \sqrt{16 + 48}}{2}\][/tex]
[tex]\[x = \frac{4 \pm \sqrt{64}}{2}\][/tex]
[tex]\[x = \frac{4 \pm 8}{2}\][/tex]
This gives us two solutions:
[tex]\[x = \frac{4 + 8}{2} = 6 \][/tex]
[tex]\[x = \frac{4 - 8}{2} = -2\][/tex]
5. Verify the Solutions:
We need to check if these solutions satisfy the original equation.
- For [tex]\(x = 6\)[/tex]:
Substitute [tex]\(x = 6\)[/tex] into the original equation:
[tex]\[\sqrt[3]{6^2 - 12} = \sqrt[3]{4 \cdot 6}\][/tex]
[tex]\[\sqrt[3]{36 - 12} = \sqrt[3]{24}\][/tex]
[tex]\[\sqrt[3]{24} = \sqrt[3]{24}\][/tex]
This is true.
- For [tex]\(x = -2\)[/tex]:
Substitute [tex]\(x = -2\)[/tex] into the original equation:
[tex]\[\sqrt[3]{(-2)^2 - 12} = \sqrt[3]{4 \cdot (-2)}\][/tex]
[tex]\[\sqrt[3]{4 - 12} = \sqrt[3]{-8}\][/tex]
[tex]\[\sqrt[3]{-8} = \sqrt[3]{-8}\][/tex]
This is true.
Since both [tex]\(x = -2\)[/tex] and [tex]\(x = 6\)[/tex] satisfy the original equation, we conclude that:
[tex]\[ \boxed{\text{Both } x = -2 \text{ and } x = 6 \text{ are true solutions.}} \][/tex]
