Please help with a few Algebra 2 questions?

You have a right triangle, so the sides have to work with the Pythagorean Theorem.

a² + b² = c²

You are given the hypotenuse to be 9, and the other two legs to be the same value "x". So substitute in "x" for a and b, and "9" for c, and solve for x:

x² + x² = 9²
2x² = 81
x² = 81 / 2
x = √(81/2)
x = 9/√2
x = (9/2)√2 (answer A)

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For the next one:

√(b - 20) + 4 = √b

To get rid of a square root, we want to square both sides. I want to get rid of the more complex one first (but the order doesn't matter as long as you don't mess up the steps, so I'll start with subtracting 4 from each side:

√(b - 20) = √b - 4

Now squaring both sides to get:

b - 20 = b - 8√b + 16

Now rearrange everything so the radical is by itself again:

-20 = -8√b + 16
-36 = -8√b
9/2 = √b

and square both sides:

b = 81/4

As a test, plug this into the original equation to make sure it's valid:

√(b - 20) + 4 = √b
√(81/4 - 20) + 4 = √81/4
√(81/4 - 80/4) + 4 = 9/2
√(1/4) + 4 = 9/2
1/2 + 4 = 9/2
1/2 + 8/2 = 9/2
9/2 = 9/2

TRUE

So x = 81/4 is the answer to the problem (answer D)