Question

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

Answer 1

A

Explanation:

So we have the equation:

`3(x-4)^(4)/(3)+16=64`

First, let's subtract 16 from both sides:

`3(x-4)^(4)/(3)=48`

Now, let's divide both sides by 3:

`(x-4)^(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!