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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!

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    $\begingroup$ Hint : Square the equation, isolate the square-root term and square again. $\endgroup$
    – Peter
    Apr 19, 2016 at 20:30
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    $\begingroup$ $6$ is the only solution. $\endgroup$
    – Peter
    Apr 19, 2016 at 20:31
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    $\begingroup$ You can check your answer with wolframalpha.com/input/?i=sqrt(x%2B19)+%2B+sqrt(x-2)+%3D+7 $\endgroup$
    – mlainz
    Apr 19, 2016 at 20:31

4 Answers 4

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Multiply both sides by $\sqrt{x+19} -\sqrt{x-2} $ to get $$x+19 -(x-2) = 7 (\sqrt{x+19} - \sqrt{x-2}) \\ 21 = 7 (\sqrt{x+19} - \sqrt{x-2}) \\ 3 =\sqrt{x+19} - \sqrt{x-2}$$

Adding this to the original equation you get $$2\sqrt{x+19}=10 \Rightarrow x+19=25 \Rightarrow x=6$$

P.S. You can find the same method employed in my answer here: to this similar question

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

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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  • $\begingroup$ As a comment, I'd imagine it is computationally easier to first divide the $32-2x$ by $2$ to get $16-x$ then square that term, but this is a trivial alteration. $\endgroup$
    – KR136
    Apr 19, 2016 at 20:58
  • $\begingroup$ Very true. I hadn't noticed this when I was first solving it. Oh well. $\endgroup$
    – zz20s
    Apr 19, 2016 at 20:59
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To solve a simple sum of radical equations, you can use with fun and profit the radical equation calculator. It gives you some steps, as describe below:

radical equation calculator

In your case, a sum of two radicals equating a constant, the "isolate one root and square" method can help: $$\sqrt{x+19}+\sqrt{x−2}=7\,.$$ So: isolate, as much as you can, one radical on one side $$\sqrt{x+19}=7 - \sqrt{x−2}$$ square: $$x+19 = 49 + x-2-14 \sqrt{x−2}$$ simplify and isolate one radical again $$28 =14 \sqrt{x−2}$$ $$2=\sqrt{x−2}$$ then square again, and you get $x=6$. Of course, you need to verify that the solution is valid, for instance that it does not involve radicals of negative numbers.

This way, you minimize products of radicals (like $\sqrt{x+19}\sqrt{x−2}$) that are easy traps. This method would work more generally, provided there are solutions, for the generic equation:

$$\sqrt{a_1 x+a_2}+\sqrt{a_3 x+a_4}=a_5$$

which you can try to exercise your style. More references:

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Hope this acceptable...

$\sqrt{{x}+\mathrm{19}}+\sqrt{{x}−\mathrm{2}}=\mathrm{7} \\ $ $\sqrt{{x}+\mathrm{19}}=\mathrm{7}−\sqrt{{x}−\mathrm{2}} \\ $ ${x}+\mathrm{19}=\mathrm{49}−\mathrm{14}\sqrt{{x}−\mathrm{2}}+\left({x}−\mathrm{2}\right) \\ $ $\mathrm{0}=\mathrm{28}−\mathrm{14}\sqrt{{x}−\mathrm{2}} \\ $ $\mathrm{2}=\sqrt{{x}−\mathrm{2}}\Rightarrow\mathrm{4}={x}−\mathrm{2}\Rightarrow{x}=\mathrm{6} \\ $

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    $\begingroup$ Please note that this question is over 5 years old and this answer doesn't add anything that isn't in another answer already. $\endgroup$
    – Magdiragdag
    Nov 11, 2021 at 8:13
  • $\begingroup$ Yes I know.., I do it for fun...😊 $\endgroup$
    – yoga mulyadi
    Nov 11, 2021 at 8:16

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