# Math Help - algebra 2 trouble....

1. ## algebra 2 trouble....

ok there are a few that im have a lot of difficulty with.... some help would be GREAT (:

1.) square root of (x + 6) - square root of (x-5) = 1

2.) square root of (2x -1) = square root of (x-1) + 1

3.) x^2/3 + x^1/3 - 2 = 0

4.) cube root of (3x-7) = cube root of (4-x)

5.) square root of (y) + fourth root of (y) = 6

i was thinking of using the "let u =" thing but im not sure if that would work...

6.) cube root of (x + 1) = sixth root of (4x + 9)

i know theres quite a few but help on any of them would be awesome <33

2. Originally Posted by eah1010
1.) square root of (x + 6) - square root of (x-5) = 1
$\sqrt{x + 6} - \sqrt{x - 5} = 1$

Isolate one of the square roots.
$\sqrt{x + 6} = \sqrt{x - 5} + 1$ <-- Square both sides

$x + 6 = \left ( \sqrt{x - 5} + 1 \right ) ^2$

$x + 6 = (x - 5) + 2 \sqrt{x - 5} + 1$ <-- Isolate the remaining square root

$2 \sqrt{x - 5} = 10$

$\sqrt{x - 5} = 5$ <-- Square again

$x - 5 = 25$

$x = 30$

Always always always check to see that this solution works in the original problem statement:
$\sqrt{30 + 6} - \sqrt{30 - 5} = 1$ (Check!)

-Dan

3. Originally Posted by eah1010
3.) x^2/3 + x^1/3 - 2 = 0
This is a quadratic equation. Let $y = x^{1/3}$. Then $y^2 = x^{2/3}$, so the original equation is
$y^2 + y - 2 = 0$

Any way you like to solve it, we get
$y = -2$ or $y = 1$

Thus
$x^{1/3} = -2$ or $x^{1/3} = 1$

Cubing both sides gives
$x = -8$ or $x = 1$

-Dan

4. Originally Posted by eah1010
6.) cube root of (x + 1) = sixth root of (4x + 9)
$\sqrt[3]{x + 1} = \sqrt[6]{4x + 9}$

For clarity express the equation this way:
$(x + 1)^{1/3} = (4x + 9)^{1/6}$

Cube both sides. The exponent on the left becomes
$\frac{1}{3} \cdot 3 = 1$

and the exponent on the right becomes
$\frac{1}{6} \cdot 3 = \frac{1}{2}$

So
$x + 1 = (4x + 9)^{1/2}$

or
$x + 1 = \sqrt{4x + 9}$

I'm sure you're sick of doing this kind of problem, but you know how to continue from here.

-Dan

5. ## (:

dan you're awesome thanks so much