#### Question

\sqrt{4-x+\dfrac{3}{4-x} }+\sqrt{x+\dfrac{3}{x} } =4

Collected in the board: Radical equations

Steven Zheng posted 4 months ago

Given

\sqrt{4-x+\dfrac{3}{4-x} }+\sqrt{x+\dfrac{3}{x} } =4
(1)

let

x = t+2
(2)

then

4-x = 4-(t+2) = 2-t
(3)

The original equation is transformed to

\sqrt{2-t+\dfrac{3}{2-t} } +\sqrt{2+t+\dfrac{3}{2+t} } =4
(4)

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
(5)

Let

p = 2-t
(6)
q = 2+t
(7)

Then,

p+q = 4
(8)

Taking square gives

p^2+q^2 = 16-2pq
(9)

pq = 4-t^2

t^2 = 4-pq
(10)

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
(11)

\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
(12)

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

Steven Zheng posted 4 months ago

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