Question

plz solve this problem ​

Answer 1

Given Question :-

Solve for x :-

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

(a) x = 2

(b) x = - 1

(c) x = 1

(d) x = 3

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

Given equation is

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

can be rewritten as after Taking lcm,

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

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

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

$\rm :\longmapsto\:x = \dfrac{1}{ \dfrac{6 - 4x - (2 - x)}{{3 - 2x }}}$

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

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

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

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

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

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

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

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

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

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

$\bf\implies \:x = 1$

• Hence, Option (c) is correct

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