quadratic equation what am I doing wrong?

Question

solve $$ \sqrt{5x+19} = \sqrt{x+7} + 2\sqrt{x-5} $$

$$ \sqrt{5x+19} = \sqrt{x+7} + 2\sqrt{x-5} \Rightarrow $$ $$ 5x+19 = (x+7) + 4\sqrt{x-5}\sqrt{x+7} + (x+5) \Rightarrow $$ $$ 3x + 17 = 4\sqrt{x-5}\sqrt{x+7} \Rightarrow $$ $$ 9x^2 + 102x + 289 = 16(x+7)(x-5) \Rightarrow $$ $$ 9x^2 + 102x + 289 = 16(x^2+ 2x - 35) \Rightarrow $$ $$ 7x^2 - 70x - 849 = 0 \Rightarrow $$ $$ b^2 - 4ac = (-70)^2 - 4 \cdot 7 \cdot (-849) = 28672=2^{12}\cdot7 $$

then I calculate the solution using the discriminant as $$ 5 + 32\frac{\sqrt7}7 $$ and $$ 5 - 32\frac{\sqrt7}7 $$

but when I plug in the values I find out that they are wrong, does it have to do with the fact that I square the equation twice? if so what is the best way to go about solving this equation?

Answer 1

You forgot a factor $4$ and wrote $x+5$ instead of $x-5$ in the third line, which should be $$ 5x+19=x+7+4\sqrt{x+7}\,\sqrt{x-5}+4(x-5) $$ giving $$ 4\sqrt{x+7}\,\sqrt{x-5}=32 $$ or $$ \sqrt{x+7}\,\sqrt{x-5}=8 $$ that becomes, after squaring, $$ x^2+2x-99=0 $$ The roots of this are $-11$ and $9$, but only the latter is a solution of the original equation, because the existence conditions on the radicals give \begin{cases} 5x+19\ge0\\[3px] x+7\ge0\\[3px] x-5\ge0 \end{cases} that is, $x\ge5$.

Answer 2

\begin{align} \sqrt{5x+19}&=\sqrt{x+7}+2\sqrt{x-5}\\ 5x+19&=x+7+4(x-5)+4\sqrt{(x+7)(x-5)}\\ 32&=4\sqrt{(x+7)(x-5)}\\ 8&=\sqrt{(x+7)(x-5)}\\ 64&=x^2+2x-35\\ 0&=x^2+2x-99 \end{align}

which gives $x=9$ and $x=-11$ are the solutions.

But both are giving something like this..

$x=9\implies \sqrt {5(9)+19}=\sqrt{9+7}+2\sqrt{9-5}$ $\implies\sqrt{64}=\sqrt{16}+2\sqrt{4}\implies 8=8$ correct know...!!!!

$x=-11\implies\sqrt {5(-11)+19}=\sqrt{-11+7}+2\sqrt{-11-5}$ $\implies\sqrt{-36}=\sqrt{-4}+2\sqrt{-16}\implies 6i=2i+4i$

But by the series of comments given by the well wishers.. i could understand $x=-11$ is not possible...