1. ## Algebra 2 problems.

Solve each equation.

1. x - 1 = √(x+2) + 3

2. 2x^2 - 6x + 5 = 0

I'm not sure how to do these?

2. Originally Posted by garbles
Solve each equation.

1. x - 1 = √(x+2) + 3

2. 2x^2 - 6x + 5 = 0

I'm not sure how to do these?
The 2nd equation is a quadratic. Use the formula to solve it.

$\displaystyle x - 1 = \sqrt{x+2} + 3$

$\displaystyle x - 4 = \sqrt{x+2}$ and $\displaystyle x\geq -2$

Now square both sides:

$\displaystyle x^2-8x+16=x+2$

You have again a quadratic. Solve it for x.

You'll get 2 solutions which are both valid.

3. Lets start by doing some rearranging:

$\displaystyle x-1=\sqrt{(x+2)}+3$ can be rearranged as

$\displaystyle x-1-3=\sqrt{x+2}$

I can then take the square of both side (to get rid of that pesky squareroot)

$\displaystyle (x-4)^{2}=(\sqrt{x+2})^{2}$

Can you take it from there?

You will not get a perfect factor and will have to use the quadratic formula to find a solution. The same holds true for your second problem.

4. Yeah I should be able to get it from there thanks.