Question

the eqiation $\sqrt{x +1} - \sqrt{x - 1} = \sqrt{4x - 1 }$ has how many solutions ?​

Answer 1

Answer:

$\huge{\bold{\mathsf{No\: solution (0 \: solutions)}}}$

Step-by-step explanation:

$\sqrt{x +1} - \sqrt{x - 1} = \sqrt{4x - 1 }$

Squaring both sides

${(\sqrt{x +1} - \sqrt{x - 1} )}^{2}={( \sqrt{4x - 1 })}^{2}$

$\implies{\mathsf{(x + 1) + (x - 1) - 2 \times \sqrt{ {x }^{2} - 1 } = 4x - 1}} \\ \\ \implies{\mathsf{2x - 4x + 1 = 2 \sqrt{ {x}^{2} - 1 } }} \\ \\\implies{\mathsf{ 1 - 2x = 2 \sqrt{ {x}^{2} - 1 } }} \\ \\ \implies{\mathsf{ \frac{1 - 2x}{2} = \sqrt{ {x}^{2} - 1 }}} \\ \\ \implies{\mathsf{ \frac{ {(1 - 2x)}^{2} }{4} = {x}^{2} - 1 }} \\ \\ \implies{\mathsf{1 + 4 {x}^{2} - 2 \times 2x = 4 {x}^{2} - 4 }} \\ \\ \implies{\mathsf{1 + 4 {x}^{2} - 4x = 4 {x}^{2} - 4 }} \\ \\ \implies{\mathsf{ 4x = 5 }} \\ \\ \implies{\boxed{\mathsf{x = \frac{5}{4} }}}$

here we get value of x

Now , put this value in given equation

$\sqrt{x +1} - \sqrt{x - 1} = \sqrt{4x - 1 }$

$\implies{\mathsf{ \sqrt{ \frac{5}{4} + 1 } - \sqrt{ \frac{5}{4} - 1 } = \sqrt{4 \times ( \frac{5}{4}) - 1} }} \\ \\ \implies{\mathsf{ \sqrt{ \frac{9}{4} } - \sqrt{ \frac{1}{4} } = \sqrt{4} }} \\ \\ \implies{\mathsf{ \frac{3}{2} - \frac{1}{2} = 2}} \\ \\ \implies{\mathsf{ \frac{2}{2} = 2 }} \\ \\ \implies{\mathsf{1 = 2}}$

Here we get L.H.S = 1 and R.H.S = 2

L.H.S ≠ R.H.S

So, this value of x doesn't satisfy the given equation

Thus , This equation have no solutions.

$\rule{200}{2}$

$\boxed{\huge{\bold{\mathsf{ IDENTITIES \: USED}}}}$ :-

(a - b )² = a² + b² - 2ab