How do I find ALL the roots

1. Feb 9, 2012

Spiralshell

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=13
(√(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
9x-2(3x+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...
9x=(32)x=(3x)2 //no idea what this is...
9x-2(3x+1)+9=0
3(x)2-6(3x)+9=0 //I don't understand how we get 3(x)2, remove X+1 for X
LET Y=3x How do we decide this?
Y2-6y+9=0 //I Understand it from here on..

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

Last edited: Feb 9, 2012
2. Feb 9, 2012

$9 = 3^2 \text{ so } 9^x = (3^2)^x = 3^{2x} = (3^x)^2$

$\text{Now } 3^{x+1} = 3\times 3^x$

$\text{So we sub them into the equation } 9^x + 2(3^{x+1}) + 9 = 0$

$(3^x)^2 + 2(3\times 3^x) + 9 = 0$

$(3^x)^2 + 6(3^x) + 9 = 0$

$\text{As you can hopefully see, this is a quadratic. We let } y = 3^x \text{ as it just makes the quadratic nice to play with and look at to find the solutions.}$

3. Feb 9, 2012

$\text{Remember the rules for indices } x^a \times x^b = x^{a+b} \text{ Try expanding the brackets, see where that takes you.}$

4. Feb 9, 2012

eumyang

Yuck. I wouldn't do that. I would rearrange the equation so that the cube root is isolated, and then cube both sides.
$\sqrt{x} - \sqrt[3]{x - 1} = 1$
$\sqrt{x} - 1 = \sqrt[3]{x - 1}$
$\left( \sqrt{x} - 1 \right)^3 = \left( \sqrt[3]{x - 1} \right)^3$
...

5. Feb 12, 2012

Spiralshell

Thanks for your hasty replies! I really appreciate your help. I am still having trouble with:

√(X)-3√(X-1)=1 //I still don't understand how to get ALL the roots, that being X=0, X=1, and X=9. How do I figure these out?

I took the advice of you both and here is what I came up with...

√(X)-1=3√(X-1)
(√(X)-1)3=X-1
(√X-1)(√X-1)
(X-√(X)-√(X)+1)(√(X)-1)
X^3/2-2X+√(X)-X+2√(X)-1
X^3/2-3X+√(X)+2√(X)-1=X-1
...
X^3/2-4X+3√(X)=0

But, now how do I find the roots?

6. Feb 12, 2012

You are almost there now. Just change it to

$x^{\frac{3}{2}} + 3\sqrt{x}=4x$

$\sqrt{x}(x+3)=4x$

Now square both sides.

7. Feb 12, 2012

epenguin

I do not understand what you are doing here

though between you and bread you seem to be getting the right result.

Natural would be to expand the two-term (binomial) cube according to the general expansion

(a - b)3 = a3 - 3a2b + 3ab2 - b3 .

Here a is √x and b is 1.

You then get an expression with √x in it and other terms, you then have to rearrange so one term involving √x is on one side of the equation and the other side contains no such terms, so that when you square both sides it will be free of square roots. (Oof, longer to explain than to do!)

8. Feb 12, 2012

eumyang

You could make a substitution: let
$y = \sqrt{x}$
Then
$x^{3/2} - 4x + 3\sqrt{x} = 0$
would become
$\text{_____} - 4\text{_____} + 3y = 0$
(I'll let you figure out what goes in the blanks.)
Solve for y, and then substitute the square root of back in.