Solve the equation $$3\sqrt{x-2}-12\sqrt{4-x}+21x-82+\sqrt{(4-x)(x-2)}=0.$$

I have tried to add $ax+b$ to each term but it's not work. Any help is appreciated. Thank you.

  • 3
    $\begingroup$ that's not an equation (there's no equals sign) $\endgroup$ – J. W. Tanner 2 days ago
  • $\begingroup$ @J.W.Tanner sorry I have edited my post $\endgroup$ – Nguyen Thy 2 days ago


Let $\sqrt{x-2}=a,\sqrt{4-x}=b$


$$21x=7(2a^2-b^2)$$ and $$82=41(a^2+b^2)$$

Replace the values to form a quadratic equation in $a$

Solve this to find $a$ in terms of $b$

  • $\begingroup$ $2a^2-b^2=3x$ not $x$ $\endgroup$ – Nguyen Thy 2 days ago
  • $\begingroup$ when it quadratic eqation in a, $\Delta$ is not in form of $k^2$. How can I solve? $\endgroup$ – Nguyen Thy 2 days ago
  • $\begingroup$ @Nguyen, what discriminant you are getting $\endgroup$ – lab bhattacharjee 2 days ago

First of all, for the square roots to be well-defined we need $2\leq x\leq 4$. Define the function:

$$f:[2,4]\to \mathbb{R},\ f(x)=21x-82-\sqrt{(4-x)(2-x)}$$

We can check that $f$ has only one root over $[2,4]$ and that is $\dfrac{51}{13}$. However this is not a solution of our equation. So let's consider the restriction that $x\in \left[2,\dfrac{51}{13}\right)\cup \left(\dfrac{51}{13},4\right]$.

Now, we can group the equation like this:




Clearly $x=\dfrac{66}{17}$ is a solution because it lies in $[2,4]$. So, it remains to discuss over:


The first fraction is always positive, while for the second we can check that $f$ is negative when $x < \dfrac{51}{13}$ and positive when $x>\dfrac{51}{13}$. Overall, the second fraction is also always positive. So no other solutions.

In conclusion $\boxed{x=\dfrac{66}{17}}$ is the unique solution.



Assuming the right side $=0$ and we are interested in real solutions,

we need $$4\ge x\ge2$$

WLOG $x-3=\cos2t,0\le t\le\pi$

$\sqrt{4-x}=\sqrt2\sin t$

$\sqrt{x-2}=\sqrt2\cos t$


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