Question

Solve for x using method of factorization

Answer 1

Given Question :-

Solve for x using method of Factorization :-

$\rm \: x = \dfrac{1}{2 - \dfrac{1}{2 - \dfrac{1}{2 - x} } }, \: \: x \: \ne \: 2$

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

Given expression is

$\rm \: x = \dfrac{1}{2 - \dfrac{1}{2 - \dfrac{1}{2 - x} } }, \: \: x \: \ne \: 2$

can be further rewritten as

$\rm \: x = \dfrac{1}{2 - \dfrac{1}{ \dfrac{2(2 - x) - 1}{2 - x} } }, \: \: x \: \ne \: 2$

$\rm \: x = \dfrac{1}{2 - \dfrac{1}{ \dfrac{4 - 2x - 1}{2 - x} } }, \: \: x \: \ne \: 2$

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

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

$\rm \: x = \dfrac{1}{\dfrac{2(3 - 2x) - 2 + x}{3 - 2x} }, \: \: x \: \ne \: 2$

$\rm \: x = \dfrac{1}{\dfrac{6 - 4x - 2 + x}{3 - 2x} }, \: \: x \: \ne \: 2$

$\rm \: x = \dfrac{1}{\dfrac{4 - 3x}{3 - 2x} }, \: \: x \: \ne \: 2$

$\rm \: x = \dfrac{3 - 2x}{4 - 3x}, \: \: x \: \ne \: 2$

$\rm \: x(4 - 3x) = 3 - 2x$

$\rm \: 4x - 3 {x}^{2} = 3 - 2x$

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

$\rm \: 3( {x}^{2} - 2x + 1) = 0$

$\rm \: {x}^{2} - 2x + 1 = 0$

$\rm \: {(x - 1)}^{2} = 0$

$\rm\implies \:x = 1, \: 1$

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

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