Question

Solve :-


$\sf x = \dfrac{1}{2 - \dfrac{1}{2 - \dfrac{1}{2 - x} } }$

Where, x ≠ 2

Topic : Quadratic Equations
Class : 10th ​

Answer 1

Answer:

x = 1

Step-by-step explanation:

Given $\sf x = \dfrac{1}{2 - \dfrac{1}{2 - \dfrac{1}{2 - x} } }$ where x ≠ 2. We need to find out the value of x.

$\sf x = \dfrac{1}{2 - \dfrac{1}{2 - \dfrac{1}{2 - x} } }$

Take (2 - x) as LCM, and solve it,

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

$\implies\:\sf x = \dfrac{1}{2 - \dfrac{1}{\dfrac{4 - 2x - 1}{2 - x}}}$

$\implies \: \sf x = \dfrac{1}{2 - \dfrac{1}{\dfrac{3- 2x}{2 - x}}}$

Now, the reciprocal of 1/(3 - 2x)/(2 - x) is (2 - x)/(3 - 2x).

$\implies \: \sf x = \dfrac{1}{2 - \dfrac{1 \times (2 - x)}{3- 2x}}$

$\implies \: \sf x = \dfrac{1}{2 - \dfrac{2 - x}{3- 2x}}$

$\implies \: \sf x = \dfrac{1}{ \dfrac{2}{1} - \dfrac{2 - x}{3- 2x}}$

LCM of 1 and (3 - 2x) is (3 - 2x). To make the denominator same multiply the 2 with (3 - 2x) and then solve the further caluations.

$\implies \: \sf x = \dfrac{1}{\dfrac{2(3 - 2x) - (2 - x)}{3- 2x}}$

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

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

$\implies \: \sf x = \dfrac{1(3 - 2x)}{4 - 3x}$

$\implies \: \sf x = \dfrac{3 - 2x}{4 - 3x}$

Corss-multiply them,

$\implies\:\sf{x(4-3x)=3-2x}$

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

$\implies\:\sf{4x-3x^{2}-3+2x=0}$

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

Take 3 as common,

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

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

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

Now, there two ways to solve it. First one is that x² - 2x + 1 is the square product of (x - 1). And second method is by splitting the middle term. So, let's proceed it!

Method 1)

→ x² - 2x + 1 = 0

→ (x - 1)² = 0

→ x - 1 = 0

→ x = 1

Method 2)

→ x² - 2x + 1 = 0

→ x² - x - x + 1 = 0

→ x(x - 1) -1(x - 1) = 0

→ (x - 1) (x - 1) = 0

→ x = 1, 1

Thus, the value of x is 1.