Math  /  Algebra

QuestionSolve: x283=2\sqrt[3]{x^{2}-8}=2 x=4x=-4 x=4x=4 x=4x=-4 or x=4x=4 no real solution

Studdy Solution

STEP 1

What is this asking? We need to find the values of xx that make this cube root equation true! Watch out! Remember that even roots have restrictions on the radicand (the stuff inside the root), but cube roots don't!
Also, remember to check your answers!

STEP 2

1. Isolate the Radical Term
2. Cube Both Sides
3. Solve the Quadratic Equation
4. Verify the Solutions

STEP 3

The radical term, x283\sqrt[3]{x^{2}-8}, is already isolated on one side of the equation, so we're good to go!
It's all set up for the next exciting step!

STEP 4

To get rid of the cube root, we're going to **cube both sides** of the equation.
Think of it like this: cubing "undoes" the cube root!
This is because n33=n\sqrt[3]{n}^3 = n for any real number nn. (x283)3=23 (\sqrt[3]{x^{2}-8})^3 = 2^3

STEP 5

This simplifies to: x28=8 x^{2}-8 = 8 Why? Because cubing a cube root leaves us with just what's inside the root!
And 22 cubed is 222=82 \cdot 2 \cdot 2 = \mathbf{8}!

STEP 6

Now, we have a nice, simple quadratic equation.
Let's **add 8** to both sides to isolate the x2x^2 term.
Remember, what we do to one side, we *must* do to the other! x28+8=8+8 x^{2}-8+8 = 8+8 x2=16 x^{2} = \mathbf{16}

STEP 7

To solve for xx, we take the **square root** of both sides.
But hold on!
Don't forget the plus-or-minus!
When we take the square root of both sides of an equation, we get two possible solutions. x2=±16 \sqrt{x^{2}} = \pm\sqrt{16} x=±4 x = \pm 4 So, our potential solutions are x=4x = \mathbf{4} and x=4x = \mathbf{-4}.

STEP 8

Let's check if our solutions actually work!
We'll substitute them back into the original equation.

STEP 9

For x=4x = \mathbf{4}: 4283=1683=83=2 \sqrt[3]{4^{2}-8} = \sqrt[3]{16-8} = \sqrt[3]{8} = 2 It works!

STEP 10

For x=4x = \mathbf{-4}: (4)283=1683=83=2 \sqrt[3]{(-4)^{2}-8} = \sqrt[3]{16-8} = \sqrt[3]{8} = 2 It works too!

STEP 11

Both x=4x = \mathbf{4} and x=4x = \mathbf{-4} are solutions to the equation!

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