I know I posted this yesterday but I really do not understand this and I did not quite understand the comment on what to do.
Find the real solutions of the equation.
x = 2 sqrt6x-36
Which do you mean?
x = 2sqrt(6x-36)
or
x = 2sqrt(6x) - 36
The first one x=2sqrt(6x-36)
To find the real solutions of the equation x = 2√(6x - 36), we need to isolate x on one side of the equation. Here's how you can solve it step by step:
Step 1: Square both sides of the equation to eliminate the square root.
(x)^2 = (2√(6x - 36))^2
Simplifying this gives:
x^2 = 4(6x - 36)
x^2 = 24x - 144
Step 2: Rearrange the equation to bring all the terms to one side, setting the equation equal to zero.
x^2 - 24x + 144 = 0
Step 3: Factorize the quadratic equation, if possible.
(x - 12)(x - 12) = 0
The equation can be factored as a perfect square.
Step 4: Set each factor equal to zero and solve for x.
x - 12 = 0
x = 12
Step 5: Check if the solution is valid by substituting it back into the original equation.
Substituting x = 12 into x = 2√(6x - 36), we get:
12 = 2√(6(12) - 36)
12 = 2√(72 - 36)
12 = 2√36
12 = 2(6)
12 = 12
The solution x = 12 is valid because it satisfies the original equation.
So, the real solution for the equation x = 2√(6x - 36) is x = 12.