Question 85363
How do I solve 2/3x = the square root of 24x - 128?
This problem, as written, is ambiguous, a few brackets would do wonders:
Do you mean:
{{{2/(3x)}}} = {{{sqrt(24x)}}} - 128
or
{{{(2/3)x}}} = {{{sqrt(24x)}}} - 128
or
{{{(2/3)x}}} = {{{sqrt(24x-128)}}}
:
The last has integer solutions, so will assume that is it:
:
Multiply equation by 3 and get rid of the denominator:
2x = 3*{{{sqrt(24x-128)}}}
:
Square both sides and you have:
4x^2 = 9(24x-128)
:
4x^2 = 216x - 1152
:
4x^2 - 216x + 1152 = 0; arrange as a quadratic equation
:
Simplify, divide equation by 4
x^2 - 54x - 288 = 0
:
Factors to:
(x - 48)(x - 6) = 0
:
Solutions are: x = +6 and x = +48
:
You should check both solutions in the original equation