First slide

Theory of equations

Question

The equation
$\sqrt{x + 1} - \sqrt{x - 1} = \sqrt{4 x - 1}, (x \in R)$

Moderate

A

no solution
B

one solution
C

two solutions
D

more than two solutions

Solution

The given equation is valid if $x + 1 \geq 0, x - 1$
≥ 0 and 4x – 1 ≥ 0 i.e. if x ≥ 1.
Squaring both the sides we get
$\begin{array}{l} x + 1 + x - 1 - 2 \sqrt{(x + 1) (x - 1)} = 4 x - 1 \\ \Rightarrow 1 - 2 x = 2 \sqrt{(x + 1) (x - 1)} \end{array}$
Squaring again, we get
$\begin{array}{l} 1 - 4 x + 4 x^{2} = 4 (x^{2} - 1) \\ \Rightarrow 4 x = 5 or x = 5 / 4 . \end{array}$
Putting this value of x in the given equation, we get
$\begin{array}{l} \sqrt{\frac{5}{4}} + 1 - \sqrt{\frac{5}{4} - 1} = \sqrt{4 (\frac{5}{4}) - 1} \\ \Rightarrow \frac{3}{2} - \frac{1}{2} = 2 or 1 = 2 \end{array}$
which is not true.
Thus, the given equation has no solution.

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