So there is this equation that I've been trying to solve but keep having trouble with.
The unit is about solving Radical equations and the question says
Solve:
$$\sqrt{x+19} + \sqrt{x-2} = 7$$
I don't want the answer blurted, I want to know how it's done, including steps please.
Thank you!
Answer
$\sqrt{x+19} + \sqrt{x-2} = 7$
Squaring both sides, we have
$x+19+2\sqrt{x+19}\sqrt{x-2}+x-2=49$
Collecting terms, we have
$2x+17+2\sqrt{x^2+17x-38}=49$
$\sqrt{x^2+17x-38}=\dfrac{32-2x}{2}$
Squaring again
$x^2+17x-38=\dfrac{1024-128+4x^2}{4}$
$x^2+17x-38=256-32x+x^2$
$49x=294$
$\therefore x=\dfrac{294}{49}=6$
We can easily verify that this is a correct solution.
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