Question

Solve the equation below. Check for extraneous solutions.

cubed root (4x-2) = 2 cubed root (x+6)

Answer 1

\[\sqrt[3]{4x-2} = 2\sqrt[3]{x+6}\]

Answer 2

Have you considered cubing both sides?

Answer 3

@tkhunny I have and I ended up with x=7, but when I plugged it back in as x in the original equation to check for extraneous solutions, I end up with a confusing result.

Answer 4

\[\sqrt[3]{4(7)-2} = 2\sqrt[3]{(7)+6} -> \sqrt[3]{26} = 2\sqrt[3]{13}\]

Answer 5

I'm not sure if the x I got was right.

Answer 6

Show me what you got after cubing both sides. Did you get 2 or 8 on the right-hand side?

Answer 7

On my first attempt I got 2. I think I forgot to also cube the 2.
On my second attempt I got 8, but my answer ended up as x=22 and when I checked my answer, I ended up with weird results.

Answer 8

4x-2 = 8(x+6)

4x-2 = 8x+48

4x-50 = 8x

-50 = 4x

x = -25/2 -- Try that.

Be more careful.

Answer 9

I see what I got wrong (I accidentally wrong 42 instead of 48).
Can you show me how to check my answer using that answer? I tried and I'm either making mistakes or I don't know how to simplify radicals.

Answer 10

\(\large \sqrt[3]{4x-2} = 2\sqrt[3]{x+6}\)

\(\large \sqrt[3]{ 4 \times \left(- \dfrac{25}{2} \right) -2} = 2\sqrt[3]{- \dfrac{25}{2} +6}\)

Answer 11

@mathstudent55 When I reduce that, I get \[\sqrt[3]{-52}=2\sqrt[3]{-13/2}\]
From there, I am stuck. Is that all I'm supposed to get? Or can I simplify it even more?

Answer 12

\(\large \sqrt[3]{ 4 \times \left(- \dfrac{25}{2} \right) -2} = 2\sqrt[3]{- \dfrac{25}{2} +6}\)

\(\large \sqrt[3]{ -50 -2} = 2\sqrt[3]{- \dfrac{25}{2} +\dfrac{12}{2}}\)

\(\large \sqrt[3]{ -52} = 2\sqrt[3]{- \dfrac{13}{2} }\)

\(\large \sqrt[3]{ -52} = \sqrt[3]{8}\sqrt[3]{- \dfrac{13}{2} }\)

\(\large \sqrt[3]{ -52} = \sqrt[3]{8 \times \left(- \dfrac{13}{2} \right)}\)

\(\large \sqrt[3]{ -52} = \sqrt[3]{- 52}\)

Answer 13

The solution works, so it's not extraneous.

Answer 14

Ohhh now I get it. Thank you so much.

Answer 15

It's also in the Domain, so it's not extraneous.