Question

The number of real solution (s) of the equation

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

is ______​

Answer 1

$\large\underline{\sf{Solution-}}$

Given equation is

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

Before we start the question, let first define the domain of the given expression.

So, the given equation is valid if

$\rm \: x + 1 \geqslant 0 \\ \rm \: x - 1 \geqslant 0 \\ 4x - 1 \geqslant 0$

So, from above, we get

$\rm \: x \geqslant 1 \: \: \: \\ \rm \: x \geqslant - 1 \\ \rm \: x \geqslant \dfrac{1}{4} \: \:$

$\bf\implies \:x \geqslant 1$

Let's solve the equation now.

Given expression is

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

On squaring both sides, we get

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

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

$\rm \: x + 1 + x - 1 - 2 \sqrt{ {x}^{2} - {1}^{2} } = 4x - 1$

$\rm \: 2x - 2 \sqrt{ {x}^{2} - 1 } = 4x - 1$

$\rm \: - 2 \sqrt{ {x}^{2} - 1 } = 4x - 1 - 2x$

$\rm \: - 2 \sqrt{ {x}^{2} - 1 } = 2x - 1$

On squaring both sides, we get

$\rm \: ( - 2 \sqrt{ {x}^{2} - 1 })^{2} = (2x - 1) ^{2}$

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

$\rm \: {4x}^{2} - 4 = {4x}^{2} + 1 - 4x$

$\rm \: - 4 = 1 - 4x$

$\rm \: - 4 - 1 = - 4x$

$\rm \: - 4x = - 5$

$\rm\implies \:x = \dfrac{5}{4}$

Verification :-

Given equation is

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

On substituting the value of x, we get

$\rm \: \sqrt{\dfrac{5}{4} + 1} - \sqrt{\dfrac{5}{4} - 1} = \sqrt{5 - 1}$

$\rm \: \sqrt{\dfrac{5 + 4}{4}} - \sqrt{\dfrac{5 - 4}{4}} = \sqrt{4}$

$\rm \: \sqrt{\dfrac{9}{4}} - \sqrt{\dfrac{1}{4}} = 2$

$\rm \: \dfrac{3}{2} - \dfrac{1}{2} = 2$

$\rm \: \dfrac{3 - 1}{2} = 2$

$\rm \: \dfrac{2}{2} = 2$

$\rm \: 1 = 2$

$\rm \: which \: is \: not \: possible$

Thus,

$\rm\implies \:\rm \: \sqrt{x + 1} - \sqrt{x - 1} = \sqrt{4x - 1} \: have \: no \: solution.$

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Additional Information

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

Discriminant, D = b² - 4ac

Answer 2

No valid solutions given

Answer above the attachment