Question

$\frac{x - 1}{x - 2} + \frac{x - 3}{x - 4} = 3 \times \frac{1}{3}$
​

Answer 1

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

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

On taking LCM, we get

$\rm :\longmapsto\:\dfrac{(x - 1)(x - 4) + (x - 3)(x - 2)}{(x - 2)(x - 4)} = \dfrac{10}{3}$

$\rm :\longmapsto\:\dfrac{ {x}^{2} - x - 4x + 4 + {x}^{2} - 3x - 2x + 6 }{ {x}^{2} - 2x - 4x + 8} = \dfrac{10}{3}$

$\rm :\longmapsto\:\dfrac{ {2x}^{2} - 10x + 10}{ {x}^{2} - 6x + 8} = \dfrac{10}{3}$

$\rm :\longmapsto\:3( {2x}^{2} - 10x + 10) = 10( {x}^{2} - 6x + 8)$

$\rm :\longmapsto\:{6x}^{2} - 30x + 30 = 10{x}^{2} - 60x + 80$

$\rm :\longmapsto\: {4x}^{2} - 30x + 50 = 0$

$\rm :\longmapsto\: {2x}^{2} - 15x + 25= 0$

$\rm :\longmapsto\: {2x}^{2} - 10x - 5x + 25= 0$

$\rm :\longmapsto\:2x(x - 5) - 5(x - 5) = 0$

$\rm :\longmapsto\:(x - 5)(2x - 5) = 0$

$\bf\implies \:x = 5 \: \: \: \: or \: \: \: \: x = \dfrac{5}{2}$

$\red{\boxed{ \bf{ \: Verification}}}$

$\green{\bf :\longmapsto\:When \: x = 5}$

Given equation is

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

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

$\rm :\longmapsto\:\dfrac{4}{3} + \dfrac{2}{1} = 3\dfrac{1}{3}$

$\rm :\longmapsto\:\dfrac{4}{3} + 2 = 3\dfrac{1}{3}$

$\rm :\longmapsto\:\dfrac{4 + 6}{3} = 3\dfrac{1}{3}$

$\rm :\longmapsto\:\dfrac{10}{3} = 3\dfrac{1}{3}$

$\rm :\longmapsto\:3\dfrac{1}{3} = 3\dfrac{1}{3}$

Hence, Verified

$\green{\bf :\longmapsto\:When \: x = \dfrac{5}{2} = 2.5 }$

Given equation is

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

$\rm :\longmapsto\:\dfrac{2.5 - 1}{2.5 - 2} + \dfrac{2.5 - 3}{2.5 - 4} = 3\dfrac{1}{3}$

$\rm :\longmapsto\:\dfrac{1.5}{0.5 } + \dfrac{ - 0.5}{ - 1.5 } = 3\dfrac{1}{3}$

$\rm :\longmapsto\:3 + \dfrac{1}{3} = 3\dfrac{1}{3}$

$\rm :\longmapsto\:3\dfrac{1}{3} = 3\dfrac{1}{3}$

Hence, Verified

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