Question

Q.4 Solve the following quadratic equation by factorization method:
1. 1/ x-2 + 2 / x-1 = 6/x​

Answer 1

Given Question :-

Solve the following equation by factorization method:

$\rm \: \dfrac{1}{x - 2} + \dfrac{2}{x - 1} = \dfrac{6}{x}$

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

Given expression is

$\rm \: \dfrac{1}{x - 2} + \dfrac{2}{x - 1} = \dfrac{6}{x}$

$\rm \: \dfrac{x - 1 + 2(x - 2)}{(x - 2)(x - 1)} = \dfrac{6}{x}$

$\rm \: \dfrac{x - 1 + 2x -4}{ {x}^{2} - 2x - x + 2 } = \dfrac{6}{x}$

$\rm \: \dfrac{3x - 5}{ {x}^{2} - 3x + 2 } = \dfrac{6}{x}$

$\rm \: 6( {x}^{2} - 3x + 2) = x(3x - 5)$

$\rm \: 6 {x}^{2} - 18x + 12 = 3 {x}^{2} - 5x$

$\rm \: 6 {x}^{2} - 18x + 12 - 3 {x}^{2} + 5x = 0$

$\rm \: 3{x}^{2} - 13x + 12 = 0$

Now, on splitting the middle terms, we get

$\rm \: 3{x}^{2} - 9x - 4x + 12 = 0$

$\rm \: 3x(x - 3) - 4(x - 3) = 0$

$\rm \: (x - 3)(3x - 4) = 0$

$\rm \: x - 3 = 0 \: \: or \: \: 3x - 4 = 0$

$\rm\implies \:x = 3 \: \: or \: \: x = \dfrac{4}{3}$

$\rule{190pt}{2pt}$

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

Answer:

Question :-

$\bigstar$ Solve the following quadratic equation by factorisation method :

$\leadsto \sf \dfrac{1}{x - 2} + \dfrac{2}{x - 1} =\: \dfrac{6}{x}$

Solution :-

$\dashrightarrow \bf \dfrac{1}{x - 2} + \dfrac{2}{x - 1} =\: \dfrac{6}{x}$

$\implies \sf \dfrac{(x - 1) + 2(x - 2)}{(x - 2)(x - 1)} =\: \dfrac{6}{x}$

$\implies \sf \dfrac{(x - 1) + 2x - 4}{x^2 - x - 2x + 2} =\: \dfrac{6}{x}$

$\implies \sf \dfrac{x - 1 + 2x - 4}{x^2 - 3x + 2} =\: \dfrac{6}{x}$

$\implies \sf \dfrac{x + 2x - 4 - 1}{x^2 - 3x + 2} =\: \dfrac{6}{x}$

$\implies \sf \dfrac{3x - 5}{x^2 - 3x + 2} =\: \dfrac{6}{x}$

By doing cross multiplication we get,

$\implies \sf 6(x^2 - 3x + 2) =\: x(3x - 5)$

$\implies \sf 6x^2 - 18x + 12 =\: 3x^2 - 5x$

$\implies \sf 6x^2 - 18x + 12 - 3x^2 + 5x =\: 0$

$\implies \sf 6x^2 - 3x^2 - 18x + 5x + 12 =\: 0$

$\implies \sf 3x^2 - 13x + 12 =\: 0$

$\implies \sf 3x^2 - (9 + 4)x + 12 =\: 0$

$\implies \sf 3x^2 - 9x - 4x + 12 =\: 0\: \: \bigg\lgroup \small \sf\bold{\pink{By\: doing\: middle-term\: break}}\bigg\rgroup$

$\implies \sf 3x(x - 3) - 4(x - 3) =\: 0$

$\implies \sf (3x - 4)(x - 3) =\: 0$

$\implies \bf 3x - 4 =\: 0$

$\implies \sf 3x =\: 4$

$\implies \sf\bold{\purple{x =\: \dfrac{4}{3}}}$

Either,

$\implies \bf x - 3 =\: 0$

$\implies \sf\bold{\purple{x =\: 3}}$

$\sf\boxed{\bold{\red{\therefore\: The\: value\: of\: x\: is\: \dfrac{4}{3}\: or\: 3\: .}}}$