Question

1. sqrt(3x+1)=3+sqrt(x-4) I know the answers for this are x=5 and 8 but I can't figure out the steps.
2. (cuberoot(x^2-3x))=cuberoot(x) I know the answers for this are x=0 and 4 but I just need the steps.

Answer 1

1) starting by squaring both sides could help

Answer 2

When I do I come out with 3x+1=9+6sqrt(x-4)+x-4 which then comes out to
3x+1 =x+5+6sqrt(x-4) and I don't know how to simply from there.

Answer 3

Well like once I do simplify it comes down to -2x+4+6sqrt(x-4) and then after that I don't know how to simplify

Answer 4

\[\sqrt{3x+1} = 3+\sqrt{x-4}\]\[\sqrt{3x+1}\sqrt{3x+1} = (3+\sqrt{x-4})(3+\sqrt{x-4})\]\[3x+1 = 3*3 + 2*3*\sqrt{x-4}+(x-4)\]\[3x+1=9+6\sqrt{x-4}+x-4\]\[3x-x=9-1-4+6\sqrt{x-4}\]\[2x=4+6\sqrt{x-4}\]

Answer 5

Now you're going to square both sides again...

Answer 6

Sorry, you'll want to move the 4 over to the left side before you square both sides...
\[2x-4 = 6\sqrt{x-4}\]Now square both sides

Answer 7

thank you so much! when you're squaring the \[6\sqrt{x-4}\] do you just say it's 36(x-4) ?

Answer 8

indeed

Answer 9

\[(6\sqrt{x-4})^2 = 6^2*(\sqrt{x-4})^2 = 36(x-4)\]

Answer 10

thank you so much!!!! I finally understand that one!!

Answer 11

I'm going to start on the other one because I'm about to call it a night.

\[\sqrt[3]{x^2-3x}=\sqrt[3]{x}\]If we raise each side to the 3rd power,
\[x^2-3x = x\]\[x^2-3x-x = x-x\]\[x^2-4x=0\]You can solve that easily, I bet...

Answer 12

Now, something very important which doesn't matter for that problem, but in general you need to watch out for whenever you solve an equation by squaring both sides: you need to take your solutions and plug them into the original formula and make sure they work! It is possible to introduce what are called extraneous solutions when you square a radical (or cube one) in the process of solving, and they will give you answers which look reasonable, but don't actually work.

Answer 13

Generally, you see these if you have a radical on one side only when you start, not if it appears on both sides. Let me see if I can make up an example for you...

Answer 14

thanks!! yeah I always forget about having to check at the end, but I will for these!! the second one was confusing me with the 0 but I factored it as (x+0)(x-4) and got it!! :)

Answer 15

Okay, here's one:
\[\sqrt{x+4}+5 = -2x\]\[\sqrt{x+4} = -2x-5\]square both sides\[x+4 = 4x^4+20x+25\]\[4x^2+19x+21 = 0\]\[x = \frac{-19\pm\sqrt{(-19)^2-4(4)(21)}}{2*4} = \frac{-19\pm\sqrt{361-336}}{8}=\frac{-19\pm\sqrt{25}}{8}\]\[x=-\frac{7}{4},x=-3\]

Now, if we try them out:
\[\sqrt{-3+4} + 5 = -2(-3)\]\[\sqrt{1}+5 = 6\]\[6=6\checkmark\]So far, so good
\[\sqrt{-\frac{7}{4}+4} + 5 = -2(-\frac{7}{4})\]\[\sqrt{\frac{9}{4}} + 5 = \frac{14}{4}\]\[\frac{3}{2}+5 = \frac{7}2\]\[5=\frac{7-3}2\]\[5=2\]Oops!
Therefore \(\large x = -\frac{7}{4}\) is an extraneous solution.

Answer 16

Yep, the second one you could just factor as \[x^2−4x=0\]\[x(x-4)=0\]\[x=0\]and
\[x-4=0\]\[x=4\]thanks to the zero product property...

Answer 17

the example with the checking is super helpful, thank you so much!! you're a lifesaver.

Answer 18

Or we could complete the square :-)

\[x^2 - 4x + (-4/2)^2 = (-4/2)^2\]\[(x-2)^2 = (-2)^2\]\[(x-2)^2 = 4\]\[x-2 = \pm2\]\[x = 2\pm2\]\[x=0,~x=4\]

Answer 19

And again, checking the answers:
the \(x=0\) one I'll leave for you :-)
\[\sqrt[3]{(4)^2-3(4)} = \sqrt[3]{4}\]\[\sqrt[3]{16-12} = \sqrt[3]{4}\]\[\sqrt[3]{4} = \sqrt[3]{4}\checkmark\]

Answer 20

yep, 0 checks out too!! thank you so much for all of your help, it was understandable and SO helpful!! :)

Answer 21

Always happy when it works out that way :-)