(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

SOLVE: √(X)-^{3}√(X-1)=1

//That is SQRT(X)-CBRT(X-1)=1; where SQRT = Square root, and CBRT = Cube root.

2. Relevant equations

None that I know of...

3. The attempt at a solution

(I would like to note I have no idea whether I am even approaching this problem the correct way.)

√(X)-^{3}√(X-1)=1

(√(X) -^{3}√(X-1))^{3}=1^{3}

(√(X)-^{3}√(X-1))((√(X)-^{3}√(X-1))((√(X)-^{3}√(X-1))

Now I foil the first two right? Then leave the 3rd and do the sum of cubes? I don't know what happens when you multiply a square root to a cube, I imagine it is something like: e.g. SQRT(X) * CBRT(X) = X^(1/2) * X(1/3) = X^(1/6) which would be^{6}√X, but then what do I do?

Also, my professor gave us the answers, but never showed us how to solve it. The answers are X=0, 1, and 9 (THIS IS NOT A GIVEN).

I feel so lost and I can't find this anywhere in my precalculus book :(! Nothing even remotely similar...

1. The problem statement, all variables and given/known data

9^{x}-2(3^{x+1})+9=0

2. Relevant equations

None that I know of...

3. The attempt at a solution

So, my professor showed us how to do this one, but I don't understand it at all...

9^{x}=(3^{2})^{x}=(3^{x})^{2}//no idea what this is...

9^{x}-2(3^{x+1})+9=0

3(^{x})^{2}-6(3^{x})+9=0 //I don't understand how we get 3(^{x})^{2}, remove X+1 for X

LET Y=3^{x}How do we decide this?

Y^{2}-6y+9=0 //I Understand it from here on..

I just need a lot of clarification to understand this stuff.

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# How do I find ALL the roots

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