2
$\begingroup$

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.

$\endgroup$
  • 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
3
$\begingroup$

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 $b$

$\endgroup$
  • $\begingroup$ $2a^2-b^2=3x$ 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
3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

Hint

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

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

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

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.