solving for x?

Question

1)
..____.............
√3x+1 + 1 = x

i don't know what i'm doing wrong but i keep ending up with 3 as the answer
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& also this question

2) if the equation x^2 - kx - 36 = 0 has x=12 as one root, what is the value of k?

Answer 1

..____.............
√3x+1 + 1 = x
..____.............
√3x+1 = x - 1

(√3x+1 )^2 = (x - 1)^2
3x + 1 = x^2 - 2x + 1
x^2 - 5x = 0
x ( x - 5 ) = 0
x = 0 x = 5

x^2 - kx - 36 = 0
x=12
x^2 - kx - 36 = 0
(x - 12)(x + 3)=0
x^2 - 9x -36 = 0
k = 9

Answer 2

In the first one, move the +1 to the other side, then square both sides. You would get 3x+1 = (x-1)^2. Simplify the right side to give 3x+1 = x^2 - 2x +1. Now combine everything on the left with everything on the right to give you a quadratic equation. (I like to have the zero on the right.) x^2 - 5x = 0. Factor out an x to give x(x-5) = 0. So x could be 0 or 5 to make this work.

If you try 0 in the original equation, it will only work if you take the negative square root of 1 (which is -1). 5 will work just fine, you get the positive square root of 16 (which is +4) plus the 1 equals 5.

In the second one, the other factor to get the -36 must be -3, so the factored quadratic would be (x+12)(x-3), and if you multiply this out, the middle factor k is +9.

Answer 3

√3x+1 + 1 = x
√3x+1 = x - 1
taking square roots on both sides
3x + 1 = (x-1)^2
3x + 1 = x^2 - 2x + 1
1 gets cancelled on both the sides. so you will end up with,
0 = x^2 - 5x
0 = x(x-5)
now u have got 2 roots
x = 0 or x = 5

substitute x = 12 in 2)
144 - 12k - 36 = 0
-12k = -108
k = 9

Answer 4

x-1=sq. rt. 3x+1

(x-1)^2=3x+1

x^2-2x+1=3x+1

x^2-2x=3x

x^2-5x=0

(x-5)(x+0)=0

x=0, x=5

2) 144-12k-36=0

-12k=-108

k=9