Solve the radical equation
$$\sqrt{4-x+\dfrac{3}{4-x} }+\sqrt{x+\dfrac{3}{x} } =4$$

Question

Solve the radical equation
$$\sqrt{4-x+\dfrac{3}{4-x}  }+\sqrt{x+\dfrac{3}{x}  }    =4$$

Uniteasy Admin · Accepted Answer

Given
$$\sqrt{4-x+\dfrac{3}{4-x} }+\sqrt{x+\dfrac{3}{x} }  =4$$n
let 
$$x = t+2$$n
then
$$4-x = 4-(t+2) = 2-t$$n
The original equation is transformed to 
$$\sqrt{2-t+\dfrac{3}{2-t} } +\sqrt{2+t+\dfrac{3}{2+t} } =4 $$n
Square both sides of the equation
$$2-t+\dfrac{3}{2-t}+2+t+\dfrac{3}{2+t}+ 2\sqrt{(2-t+\dfrac{3}{2-t})(2+t+\dfrac{3}{2+t}) } =16$$
Simplify
$$\dfrac{3}{2-t}+\dfrac{3}{2+t}+ 2\sqrt{(2-t+\dfrac{3}{2-t})(2+t+\dfrac{3}{2+t}) } =12$$n
Let
$$p = 2-t$$n
$$q = 2+t$$n
Then,
$$p+q = 4$$n
Taking square gives
$$p^2+q^2 = 16-2pq$$n
$$pq = 4-t^2$$
$$t^2 = 4-pq$$n
Equation (4) is transformed to
$$\dfrac{3}{p}+\dfrac{3}{q}+2\sqrt{(p+\dfrac{3}{p} )(q+\dfrac{3}{q} )} =12  $$ 
$$ \dfrac{3(p+q)}{pq} +2\sqrt{pq+\dfrac{3p}{q}+\dfrac{3q}{p}+\dfrac{9}{pq} } = 12 $$
$$ \dfrac{12}{pq} +2\sqrt{pq+\dfrac{3(p^2+q^2)}{pq} +\dfrac{9}{pq} } = 12 $$n
$$\sqrt{pq+\dfrac{3(p^2+q^2)}{pq} +\dfrac{9}{pq} } = 6(1-\dfrac{1}{pq} )$$
Taking square on both sides andsubstituring (8) gives
$$pq+\dfrac{3(16-2pq)}{pq} +\dfrac{9}{pq} =6^2(1-\dfrac{1}{pq} )^2$$
Let
$$m = pq$$n
Then
$$m+\dfrac{3(16-2m)}{m} +\dfrac{9}{m} =36(1-\dfrac{1}{m} )^2$$
Multiply $$m^2 $$on both sides
$$m^3+48m-6m^2+9m =36(m^2-2m+1)$$
$$m^3-42m^2+129m-36=0$$
Factorizing gives
$$(m-3)(m^2-39+12) = 0$$
Solving for $$m$$ gives three real roots
$$\begin{cases} m_1 =3 \ m_2 = \dfrac{39+\sqrt{1473} }{2} \approx 38.689841062395 
\ m_3 = \dfrac{39-\sqrt{1473} }{2} \approx 0.31015893760452 
\end{cases}$$
Since $$m = pq$$, substituting to (9) will result in $$t^2$$
When $$m = 3$$,
$$t^2 = 4-pq = 1$$
then $$t = 1$$ or $$t = -1$$
Substituting the value of $$t$$ to (2) gives
$$x = 1$$ or $$x = 3$$
Substituting the values of $$x = 1$$ or $$x = 3$$ to (1) respectively to verify shows $$x = 1$$ or $$x = 3$$ are indeed the solutions to hold the equation true.
When $$m = \dfrac{39+\sqrt{1473} }{2} \approx 38.689841062395 $$,
$$t^2 = 4-pq < 0 $$, $$t$$ is complex number. If $$ t$$ is complex number, the Radicants under two square roots are not conjugate. So addition after taking square root will not result in real number. So $$m_2$$ is not the solution.
 $$\sqrt{2-t+\dfrac{3}{2-t} } +\sqrt{2+t+\dfrac{3}{2+t} } =4 $$
When $$m = \dfrac{39-\sqrt{1473} }{2} \approx 0.31015893760452 $$
$$t^2 = 4-0.31015893760452\approx 3.689841062$$
$$t = \pm 1.920895901$$
Substituting the value of $$t$$ to (2) gives
$$x = 3.920895901 $$ or $$x = 0.079104099$$
Substituting the values to (1) shows they are not holding the equation true.
In summary, there are two solutions for the radical equation
$$x = 1$$ or $$x = 3$$

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