If \(x \in \mathbf{R}\), then one of the solutions of \(\sqrt{x+1}-|\sqrt{x-1}|=\sqrt{4 x-1}\) among the…
If \(x \in \mathbf{R}\), then one of the solutions of \(\sqrt{x+1}-|\sqrt{x-1}|=\sqrt{4 x-1}\) among the following is
- \(x=\frac{5}{4}\)
- \(x=\frac{-5}{4}\)
- \(x=0\)
- \(x=1\)
Solution
\(\begin{aligned}
& \sqrt{x+1}-|\sqrt{x-1}|=\sqrt{4 x-1} \\
& \Rightarrow \sqrt{x+1}-\sqrt{4 x-1}=|\sqrt{x-1}|
\end{aligned}\)
Squaring on both sides
\(\begin{array}{ll}
\Rightarrow & x+1+4 x-1-2 \sqrt{(x+1)(4 x-1)}=x-1 \\
\Rightarrow & 5 x-2 \sqrt{(x+1)(4 x-1)}=x-1 \\
\Rightarrow & 4 x+1=2 \sqrt{(x+1)(4 x-1)}
\end{array}\)
Squaring on both sides
\(\begin{array}{lll}
\Rightarrow & 16 x^2+1+8 x=4\left(4 x^2+3 x-1\right) \\
\Rightarrow & 16 x^2+8 x+1=16 x^2+12 x-4 \\
\Rightarrow & 4 x=5 \Rightarrow x=\frac{5}{4}
\end{array}\)
Asked in: AP EAMCET 2020 (17 Sep Shift 1)
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