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

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.

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

Hint

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

$$a^2+b^2=2,2a^2-b^2=3x$$

$$21x=7(2a^2-b^2)$$ 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 $$b$$

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

$$3(\sqrt{x-2}-4\sqrt{4-x})+\left[21x-82+\sqrt{(4-x)(x-2)}\right]=0$$

or

$$\frac{3(17x-66)}{\sqrt{x-2}+4\sqrt{4-x}}+\frac{2(17x-66)(13x-51)}{21x-82-\sqrt{(4-x)(x-2)}}=0$$

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

$$\frac{3}{\sqrt{x-2}+4\sqrt{4-x}}+\frac{2(13x-51)}{21x-82-\sqrt{(4-x)(x-2)}}=0$$

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.

Hint

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

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

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

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

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