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Solve in $\Bbb R$ the following equation: $\displaystyle\sqrt{x+4-4\sqrt {x}}+\sqrt{x+9-6\sqrt {x}}=1$.$\qquad\text{(1)}$

My try: The set of Perfect squares is: $\left\{\color{blue}{0,1,4,9,16,25,36,49\ldots}\right\}$

Some guesswork: $$\begin{align} \sqrt{\color{blue}0+4-4\sqrt {\color{blue}0}}=2 & & \sqrt{\color{blue}0+9-6\sqrt {\color{blue}0}}=3 && \text{$\boldsymbol\times$} \\\,\\ \sqrt{\color{blue}1+4-4\sqrt {\color{blue}1}}=1 & & \sqrt{\color{blue}1+9-6\sqrt {\color{blue}1}}=2 && \text{$\boldsymbol\times$}\\\,\\ \sqrt{\color{blue}4+4-4\sqrt {\color{blue}4}}=0 & & \sqrt{\color{blue}4+9-6\sqrt {\color{blue}4}}=1 &&\text{$\boldsymbol\checkmark$} \end{align}$$

So the solution is $x=4$.

But is there a more direct way, i.e. to find $x$ without having to do guesswork?

Thanks in advance.

  • 1
    Keep squaring both sides until you can solve for x? – zerosofthezeta Dec 28 '13 at 19:32
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    @zeta That approach will almost never work. – mathematics2x2life Dec 28 '13 at 19:41
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    @mathematics2x2life: I agree with mathematics2x2life: If you call the first radical $a$ and the second $b$, then the problem is $a+b=1$. If you move $b$ to one side to get rid of one radical, we get $a=(1-b)$ or $a^2 = 1 - 2 b + b^2$. Move things around to get $2b = 1 + b^2 - a^2$. Square again you end up with $0=0$! Ugh – user44197 Dec 28 '13 at 20:15
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    Could one tell me where I am wrong ? I find that this equation has an infinite number of solutions, all of them being between x=4 and x=9. Thanks. – Claude Leibovici Dec 29 '13 at 05:28
  • @ClaudeLeibovici Nope, that's wrong. –  Dec 29 '13 at 09:48
  • @Adobe. Could you clarify, please ? Happy New Year. – Claude Leibovici Dec 29 '13 at 10:17
  • @ClaudeLeibovici I tried $x=9$ and it doesn't work, let me do more work. Happy New Year! –  Dec 29 '13 at 13:20
  • @ClaudeLeibovici I don't know what's happening when I plug $x=8.6$ in Wolfram|Alpha since I get $1.000000...$ And I don't know if it is just an error by WA. –  Dec 30 '13 at 17:50
  • @Adobe.This is what I was telling you. The solutions cover an infinite range between 4 and 9. May I suggest you plot the function ? Happy New Year !! – Claude Leibovici Dec 31 '13 at 04:48
  • @ClaudeLeibovici You're right! Happy New Year!! :) –  Dec 31 '13 at 09:30

2 Answers2

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Hint:

$$x+4-4\sqrt {x}=(\sqrt{x}-2)^2$$

$$x+9-6\sqrt {x}=(\sqrt{x}-3)^2$$

And don't forget the absolute values. ;)

The simplest solution is to break the problem in $3$ cases:$\sqrt{x} \lt2$ or $2 \leqslant \sqrt{x} \leqslant 3$ or $3\lt \sqrt{x}$.

N. S.
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Just to expand on @zeta's comment, because it is possible to square in order to eliminate radicals:

$$\left(\sqrt{x+4-4\sqrt{x}}\right)^2=\left(1-\sqrt{x+9-6\sqrt{x}}\right)^2$$gives, on simplifying: $$2\sqrt{x}-6=-2\sqrt{x+9-6\sqrt x}$$ cancelling the factor $2$ and squaring gives $0=0$.

The problem is that squaring allows solutions for both signs of the square root (and therefore includes unwanted solutions, which have to be eliminated by studying cases). If the solution set consists of isolated points, this can be a way of proceeding.

It is also important to check the validity of the original formula for the solutions once they are found. Here the formula makes no sense over $\mathbb R$ for negative values of $x$, yet the squaring method does not exclude negative $x$.

[This is really a comment, but was too long for that]

Mark Bennet
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  • I agree. As I had mentioned, you end up with $0=0$. – user44197 Dec 28 '13 at 20:56
  • @user44197 I didn't see your comment before I wrote mine - but we definitely agree. I was interested because the solution to the question is an interval, and yet normally squaring systematically to remove the roots leaves a polynomial. It was predictable that squaring would give an identity - but in many cases, contra one comment - squaring will succeed - it is just messy and introduces spurious roots. – Mark Bennet Dec 28 '13 at 22:52