Question

The number of real roots of the equation
\[ \sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}=\frac{13}{6} \] is

• A. \(1\)
• B. \(2\)
• C. \(3\)
• D. \(4\)

Hint:

For expressions of the form \(\sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}\), substitute \(t=\sqrt{\frac{x}{1-x}}\). Then the second term becomes \(\frac{1}{t}\).

Answer 1

The Correct Option is B

Step 1: Find the domain.
For the square roots to be real, both fractions must be positive.
So,
\[ \frac{x}{1-x}\gt 0 \]
This gives
\[ 0\lt x\lt 1 \]

Step 2: Substitute a variable.
Let
\[ t=\sqrt{\frac{x}{1-x}} \]
Since \(0\lt x\lt 1\), we have
\[ t\gt 0 \]
Also,
\[ \sqrt{\frac{1-x}{x}}=\frac{1}{t} \]
Therefore, the equation becomes
\[ t+\frac{1}{t}=\frac{13}{6} \]

Step 3: Convert into quadratic equation.
Multiplying by \(6t\), we get
\[ 6t^2+6=13t \]
\[ 6t^2-13t+6=0 \]

Step 4: Solve the quadratic equation.
\[ 6t^2-13t+6=0 \]
\[ 6t^2-9t-4t+6=0 \]
\[ 3t(2t-3)-2(2t-3)=0 \]
\[ (3t-2)(2t-3)=0 \]
Thus,
\[ t=\frac{2}{3} \] or
\[ t=\frac{3}{2} \]
Both values are positive, so both are valid.

Step 5: Count the real roots.
For each positive value of \(t\), there is exactly one value of \(x\) because
\[ t^2=\frac{x}{1-x} \]
gives a unique \(x\in(0,1)\).
Hence, the number of real roots is
\[ 2 \]

Step 6: Final conclusion.
Therefore,
\[ \boxed{2} \]