Help with some Algebra 2 problems please?? (Radicals)?

Question

√(1+8v) - 2 = v
√(x-6) = 3 + √x

*everything in the parenthesis is in the radical

Answer 1

√(1 + 8v) - 2 = v
√(1 + 8v) = v + 2
1 + 8v = v² + 4v + 4
v² - 4v + 3 = 0
(v-3)(v-1) = 0
v = 3 or v = 1
checking, since squaring both sides can introduce extraneous roots,
√(1 + 8•3) - 2 =
√25 - 2 =
5 - 2 = 3 ....... ok
√(1 + 8•1) - 2 =
√9 - 2 =
3 - 2 = 1 ........ also ok.

√(x-6) = 3 + √x
x - 6 = 9 + 6√x + x
-15 = 6√x
-5/2 = √x
this has no solution if x is real.

Answer 2

√(1+8v) - 2 = v
√(1+8v) = v+2
1+ 8v = v^2 + 4v + 4
v^2 - 4v + 3 = 0
(v-3)(v-1) = 0
v = 3,1

Check ( there is usually an extraneous solution when you square both sides)
√(1+8(1)) - 2 = 1
√(9) - 2 = 1
3 - 2 = 1
1 = 1

√(1+8(3)) - 2 = 3
√(1+24) - 2 = 3
√(25) - 2 = 3
5 - 2 = 3
3=3
since this is also true, there are no extraneous solutions, both of the are true...


√(x-6) = 3 + √x
x-6 = 9 + 6√x + x
-15 = 6√x
-5/2 = √x

since a square root cannot equal a negative number, there is no solution...

Answer 3

sqrt(1+8v) = v+2

square both sides:
1+8v = v^2 + 4v + 4
v^2 -4v + 3 = 0
(v-3)(v-1) = 0
v = 1, 3


sqrt(x-6) = 3 + sqrt(x)

square both sides:
x-6 = 9 + 6sqrt(x) + x
6 sqrt(x) = -15
sqrt(x) = -5/2
x = 25/4

Hope this helps! :)

Answer 4

square everything in both equations, then get everything over to one side, and then factor. :D

1+8v=(v+2)^2
0=v^2-4v+3
0=(v-3)(v-1)