square root (x+6) +1 = square root (2x-11)?

Question

please help I am having a hard time trying to figure this out.

Answer 1

square both sides
x+6+1=2x-11
7+11=2x-
18=x

x=18

Answer 2

I'll assume
[sqrt(x+6)] +1 = sqrt(2x-11)

[(x+6)/sqrt(x+6)] + [sqrt(x+6)/sqrt(x+6)] = sqrt(2x-11)

[ x + 6 + sqrt(x+6) ] / sqrt(x+6) = sqrt(2x-11)

NOW SQUARING EVERYTHING

[ x + 6 + sqrt(x+6) ]^2 / sqrt(x+6)^2 = sqrt(2x-11)^2

[[ x^2 + 6x + x[sqrt(x+6)] + 6x + 36 + 6[sqrt(x+6)] + x[sqrt(x+6)] + 6[sqrt(x+6)] + x + 6 ]] / (x+6) = 2x-11

x^2 + 13x + 42 + 2x[sqrt(x+6)] + 12[sqrt(x+6)] = (2x-11)(x+6)
x^2 + 13x + 42 + 2x[sqrt(x+6)] + 12[sqrt(x+6)] = 2x^2+x-66
2x[sqrt(x+6)] + 12[sqrt(x+6)] = x^2 - 12x - 108
(2x+12) [sqrt(x+6)] = x^2 - 12x - 108
[sqrt(x+6)] = (x^2-12x-108) / (2x+12) <---divide one into the other

[sqrt(x+6)] = (1/2)x - 9 <--square both sides again
x + 6 = (1/4)x^2 - 9x + 81
(1/4)x^2 - 10x + 75 = 0 <---from here use the quadratic equation

[[ 10 +/- sqrt [ 100 - 4*(1/4)*75] ]]/ 2*(1/4)
[10 +/- sqrt 25] / (1/2)
2* (10 +/- 5)
x = 30 or 10 but checking in original equation exempts 10 so
x = 30 <-----ANSWER

Answer 3

ok...

square everything:

[(x+6)^1/2 + 1]^2 = 2x-11
(x+6) + 2(x+6)^1/2 + 1 = 2x - 11
2(x+6)^1/2 = x - 18
(x+6)^1/2 = 1/2x - 9

Now square again...

x+6 = 1/4x^2 - 9x + 81

solve for zero:

1/4x^2 - 10x + 75 = 0
x^2 - 40x + 150 = 0

then use the quadratic formula to solve

Hope this helps!

Answer 4

is 1 outside of the radical?