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.
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.