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Math  /  Algebra

QuestionWhich statement is true about the solution of x2−123=4x3\sqrt[3]{x^{2}-12}=\sqrt[3]{4 x} ? x=−2x=-2 is an extraneous solution, and x=6x=6 is a true solution. x=6x=6 is an extraneous solution, and x=−2x=-2 is a true solution. Both x=−2x=-2 and x=6x=6 are extraneous solutions. Both x=−2x=-2 and x=6x=6 are true solutions.

Studdy Solution

STEP 1

1. The equation involves cube roots, which are defined for all real numbers.
2. We need to solve the equation and check each potential solution to determine if it is true or extraneous.

STEP 2

1. Set the cube root expressions equal to each other.
2. Eliminate the cube roots by cubing both sides.
3. Solve the resulting polynomial equation.
4. Check each solution to determine if it is true or extraneous.

STEP 3

Set the cube root expressions equal to each other:
x2−123=4x3 \sqrt[3]{x^2 - 12} = \sqrt[3]{4x}

STEP 4

Eliminate the cube roots by cubing both sides of the equation:
(x2−123)3=(4x3)3 (\sqrt[3]{x^2 - 12})^3 = (\sqrt[3]{4x})^3
This simplifies to:
x2−12=4x x^2 - 12 = 4x

STEP 5

Rearrange the equation to form a standard quadratic equation:
x2−4x−12=0 x^2 - 4x - 12 = 0

STEP 6

Solve the quadratic equation using the quadratic formula:
The quadratic formula is given by:
x=−b±b2−4ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
For our equation, a=1 a = 1 , b=−4 b = -4 , and c=−12 c = -12 .
x=−(−4)±(−4)2−4⋅1⋅(−12)2⋅1 x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-12)}}{2 \cdot 1} x=4±16+482 x = \frac{4 \pm \sqrt{16 + 48}}{2} x=4±642 x = \frac{4 \pm \sqrt{64}}{2} x=4±82 x = \frac{4 \pm 8}{2}
This gives the solutions:
x=4+82=6 x = \frac{4 + 8}{2} = 6 x=4−82=−2 x = \frac{4 - 8}{2} = -2

STEP 7

Check each solution in the original equation to determine if it is true or extraneous.
First, check x=6 x = 6 :
62−123=4⋅63 \sqrt[3]{6^2 - 12} = \sqrt[3]{4 \cdot 6} 36−123=243 \sqrt[3]{36 - 12} = \sqrt[3]{24} 243=243 \sqrt[3]{24} = \sqrt[3]{24}
This is true.
Next, check x=−2 x = -2 :
(−2)2−123=4⋅(−2)3 \sqrt[3]{(-2)^2 - 12} = \sqrt[3]{4 \cdot (-2)} 4−123=−83 \sqrt[3]{4 - 12} = \sqrt[3]{-8} −83=−83 \sqrt[3]{-8} = \sqrt[3]{-8}
This is also true.
Both x=−2 x = -2 and x=6 x = 6 are true solutions. Therefore, the correct statement is:
Both x=−2 x = -2 and x=6 x = 6 are true solutions.

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