Question

Please help me :)) algebra 2! what are the real roots?

Answer 1

Answer:

A

Step-by-step explanation:

So we have the equation:

$3(x-4)^\frac{4}{3}+16=64$

First, let's subtract 16 from both sides:

$3(x-4)^\frac{4}{3}=48$

Now, let's divide both sides by 3:

$(x-4)^\frac{4}{3}=16$

Remember that with fractional exponents, we can move the denominator into the root position. Therefore:

$(\sqrt[3]{x-4})^4=16$

Let's take the fourth root of both sides. Since we're taking an even root, make sure to have the plus-minus symbol!

$\sqrt[3]{x-4} =\pm 2$

Cube both sides. Since we're cubing, the plus-minus stays.

$x-4=\pm 8$

Add 4 to both sides.

$x=\pm 8+4$

Calculator:

$x=8+4\text{ or } x=-8+4\\x=12\text{ or } x=-4$

So, our answer is A.

And we're done!