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I recognize the two difference of squares: $49-x^2$ and $25-x^2$.

I squared the equation to get: ${49-x^2}-2(\sqrt{(49-x^2)(25-x^2)})+{25-x^2}=9$

However, I can't quite figure out how to remove the root in the middle. Any help is appreciated.

Kohei Sanno
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    Hint: Rearrange the terms and square again. – Luca Bressan Jul 29 '18 at 13:47
  • You could move the square root to the RHS and gather other terms to the LHS. Then taking square and the root is removed. – xbh Jul 29 '18 at 13:48
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    Bring everything else except the root to the R.H.S. and square the equation again. After this step there should be only expressions involving $x^4$ and $x^2$ and numbers which is a biquadratic equation. This type you can solve by setting $z=x^2$ and afterwords you need to check every single solution because to square is not a aquivalent reshape operation. – mrtaurho Jul 29 '18 at 13:48
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    If you multiply by $\sqrt{49-x^2}+\sqrt{25-x^2}$ you get $\sqrt{49-x^2}+\sqrt{25-x^2}=\frac{49-25}{3}$. If you call $A=\sqrt{49-x^2}$ and $B=\sqrt{25-x^2}$, you just got a system of equations $A-B=3$, $A+B=\frac{49-25}{3}$. So, $A=\frac{3+\frac{49-25}{3}}{2}$, from where you get $x^2=49-\left(\frac{3+\frac{49-25}{3}}{2}\right)^2$ –  Jul 29 '18 at 13:54

7 Answers7

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Hint. You may just write $$ {49-x^2}+{25-x^2}-9=2\sqrt{(49-x^2)(25-x^2)} $$ simplify and square both sides once again.

Olivier Oloa
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Another solution \begin{align*} \sqrt {49 - x^2} - \sqrt {25 - x^2} &= 3\\ \sqrt {49-x^2} &= 3+ \sqrt {25-x^2}\\ 49-x^2 &= 25 -x^2 + 9 + 6\sqrt {25-x^2}\\ 15 &= 6 \sqrt {25 - x^2}\\ 25 - x^2 &= \frac {25} 4\\ x^2 &= \frac {75}4\\ x &= \pm \frac 5 2 \sqrt 3. \end{align*}

xbh
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We have $$\sqrt{49-x^2}-\sqrt{25-x^2}=3$$ and $$\left(\sqrt{49-x^2}-\sqrt{25-x^2}\right)\left( \sqrt{49-x^2}+\sqrt{25-x^2}\right)=3\left(\sqrt{49-x^2}+\sqrt{25-x^2}\right),$$ which gives $$\sqrt{49-x^2}+\sqrt{25-x^2}=8$$ and from here $$\sqrt{49-x^2}=5.5,$$ which gives the answer: $$\left\{-\sqrt{18.75},\sqrt{18.75}\right\}.$$

Michael Rozenberg
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First express one root: $$\sqrt{49-x^2}=3+\sqrt{25-x^2}$$ Now square:

$$ 49-x^2=9+6\sqrt{25-x^2}+ 25-x^2\implies \boxed{5=2\sqrt{25-x^2}}$$

and square again...

nonuser
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Note that $$\sqrt{49-x^2}+\sqrt{25-x^2}=\frac{(49-x^2)-(25-x^2)}{\sqrt{49-x^2}-\sqrt{25-x^2}}=\frac{24}{3}=8\,.$$ Together with $\sqrt{49-x^2}-\sqrt{25-x^2}=3$, we conclude that $$\sqrt{49-x^2}=\frac{3+8}{2}=\frac{11}{2}\,.$$

Therefore, $x^2=\dfrac{75}{4}$, or $x=\pm\dfrac{5\sqrt{3}}{2}$.

P.S.: Oopsie, nextpuzzle already made a comment, giving the same solution.

Batominovski
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This is a good start. Now rearrange the equation so that only the square root remains on the left hand side. Take square again. Then you obtain a bi-quadratic equations, or "fake quartic" equation. Introduce a new variable $y=x^2$, and then you have a quadratic equation in $y$ that you can solve easily.

Note that taking squares is not an equivalent transformation, so in the end, you have to check all the solutions. Some, or in fact, all of them can be extraneous.

A. Pongrácz
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Set $\;\;t^2=25-x^2,\;$ the equation becomes $$\begin{align}\sqrt{24+t^2}-\sqrt{t^2}&=3\\ \sqrt{24+t^2}&=|t|+3\\24+t^2&=t^2+6|t|+9\\|t|&=\frac{5}{2} \end{align}$$ Then $x^2=25-\frac{25}{4}=3\cdot \frac{25}{4},$ from where $x=\pm \frac{5}{2}\sqrt 3.$

user376343
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