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?

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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).

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Are you sure that $-\frac{77}{4}$ is correct? –  Miroslav Cetojevic May 15 '12 at 15:36
@MiroslavCetojevic oops, you are right. I edited the answer... –  Thomas May 15 '12 at 15:39
I got $-\frac{133}{4}$, but otherwise it looks like this was the hint I needed. Thanks! –  Miroslav Cetojevic May 15 '12 at 15:42
@MiroslavCetojevic, again: you are right. I am not good with numbers :) Glad that it helped though. –  Thomas May 15 '12 at 15:43

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).

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Suggestion:

1. Multiply the given equation by $4$ to have integer coefficients only. (You may omit this step)
2. Expand the LHS. $$4x+24\sqrt{x}-28=105.$$
3. Isolate the term with a square root on one side. $$24\sqrt{x}=133-4x.$$
4. 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. $$x_1=\frac{49}{4}.$$
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