Question

Solve for x
$\sf \: \sqrt{2x - 1} + \sqrt{3x - 2} = \sqrt{4x - 3} + \sqrt{5x - 4}$
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Class NTSE Quadratic Equations.

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Answer 1

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

Given equation is

$\rm \: \sqrt{2x - 1} + \sqrt{3x - 2} = \sqrt{4x - 3} + \sqrt{5x - 4}$

To solve such type of equations,

Let assume that

$\red{\rm :\longmapsto\:a = \sqrt{2x - 1}}$

$\red{\rm :\longmapsto\:b = \sqrt{3x - 2}}$

$\red{\rm :\longmapsto\:c = \sqrt{4x - 3}}$

$\red{\rm :\longmapsto\:d = \sqrt{5x - 4}}$

So, given equation can be rewritten as

$\rm :\longmapsto\:a + b = c + d - - - (1)$

Now, Consider

$\red{\rm :\longmapsto\: {a}^{2} - {b}^{2}}$

$\rm \:  =  \:2x - 1 - (3x - 2)$

$\rm \:  =  \:2x - 1 - 3x + 2$

$\rm \:  =  \:1 - x$

$\red{\rm :\longmapsto\: \boxed{ \rm \:{a}^{2} - {b}^{2} = 1 - x } - - - - - (2)}$

Now, Consider

$\red{\rm :\longmapsto\: {c}^{2} - {d}^{2}}$

$\rm \:  =  \:4x - 3 - (5x - 4)$

$\rm \:  =  \:4x - 3 - 5x + 4$

$\rm \:  =  \:1 - x$

$\red{\rm :\longmapsto\: \boxed{ \rm \:{c}^{2} - {d}^{2} = 1 - x } - - - - - (3)}$

From equation (2) and (3), we concluded that

$\rm :\longmapsto\: {a}^{2} - {b}^{2} = {c}^{2} - {d}^{2}$

$\rm :\longmapsto\:(a + b)(a - b) = (c + d)(c - d)$

So, using equation (1), we get

$\rm :\longmapsto\:(c + d)(a - b) = (c + d)(c - d)$

$\rm :\longmapsto\:a - b = c - d - - - - (4)$

On adding equation (1) and equation (4), we get

$\rm :\longmapsto\:2a = 2c$

$\rm :\longmapsto\:a = c$

$\rm :\longmapsto\: \sqrt{2x - 1} = \sqrt{4x - 3}$

On squaring both sides,

$\rm :\longmapsto\:2x - 1 = 4x - 3$

$\rm :\longmapsto\:2x -4x= 1- 3$

$\rm :\longmapsto\: -2x= - 2$

$\bf :\longmapsto\:x=1$

Verification

Given equation is

$\rm \: \sqrt{2x - 1} + \sqrt{3x - 2} = \sqrt{4x - 3} + \sqrt{5x - 4}$

On substituting the value of x = 1

$\rm \: \sqrt{2 - 1} + \sqrt{3 - 2} = \sqrt{4 - 3} + \sqrt{5 - 4}$

$\rm \: \sqrt{1} + \sqrt{1} = \sqrt{1} + \sqrt{1}$

$\rm \: 1 + 1 = 1 + 1$

$\rm \: 2 = 2$