Root equation - What am I missing? There's a problem of which I know the solution but not the solving process:

$(\sqrt{x} + 7)(\sqrt{x} - 1) = \frac{105}{4}$

I'm convinced that up to:

$x + 6\sqrt{x} - 7 = \frac{105}{4}$

everything is correct.
But afterwards, I never seem to be able to get to the solution of $x = \frac{49}{4}$
How can this equation be solved?
 A: Hint: Your equation can be rewritten as
$$
(\sqrt{x})^2 + 6 \sqrt{x} -\frac{133}{4} = 0
$$
And so it is a quadratic equation in $\sqrt{x}$ that you can solve. So you find that $\sqrt{x}$ is equal to some number(s).
A: Hint $\ $ Put $\:z = \sqrt{x},\:$ clear denominators and rearrange to obtain the quadratic equation$\:0 = 4 z^2 + 24 z - 133 = (2z-7)(2z+19),\:$ solvable by the quadratic formula, or the Rational Root Test or  AC method, etc.
Alternatively, purely arithmetically, if it has rational root $\:\sqrt{x} = a/b\:$ in lowest terms then
$$ (\sqrt{x} +7)(\sqrt{x}-1)\ =\ \frac{a+7b}b\ \frac{a-b}{b}\ =\ \frac{3\cdot 5\cdot 7}4 $$
Since $\:b\:$ is coprime to $\:a,\:$ also $\:b\:$ is coprime to $\:a+7b,\ a-b,\:$ so we deduce  $\:b = 2.\:$  Hence
$$  (a + 14)\:(a-2)\ =\ 3\cdot 5\cdot 7 $$
So we seek a factorization of $\:3\cdot 5\cdot 7\:$ whose factors differ by $\:16 = a+14-(a-2).\:$ Checking the few possible factorizations we find $\:21\cdot 5\:\Rightarrow\:a=7,\:$ and $\:-5\:(-21)\:\Rightarrow\:a = -19.\:$ Notice that this solution depends on uniqueness of factorization (as does RRT and the AC method).
A: Suggestion:


*

*Multiply the given equation by $4$ to have integer coefficients only. (You may omit this step)

*Expand the LHS.
$$4x+24\sqrt{x}-28=105.$$

*Isolate the term with a square root on one side.
$$24\sqrt{x}=133-4x.$$

*Square both sides (this may introduce a new solution) and simplify. You get a quadratic equation. One of its two solutions satisfies the original equation. $$x_1=\frac{49}{4}.$$

