Question

find the value of x./solved
$x = \frac{1}{2 - \frac{1}{2 - \frac{1}{2 - x} } }$
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Answer 1

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

Given equation is

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

can be rewritten as

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\rm \: (x - 1)(x - 1) = 0$

$\bf\implies \:x = 1$

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

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

Answer 2

Answer:

x = 1

$\rule{70mm}{2pt}$

Step-by-step explanation:

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

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

Make the denominator same or just simply take (2 - x) as L.C.M.

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

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

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

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

Again take the L.C.M. or make the denominator same. To make the denominator same multiply and divide the 2/1 with (3 - 2x) and then solve the further calculations.

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

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

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

Cross-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 further. 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)

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

$\implies\sf{(x - 1)^2 = 0\:\:Used\: identity:\:(a-b)^2}$

$\implies\sf{x - 1 = 0}$

$\implies\sf{x = 1}$

Method 2)

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

The above equation is in the standard form of ax² + bx + c = 0 (quadratic equation). Now, solve it by factorisation. Factor the expression and then set each of the factors to zero.

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

$\implies\sf{x(x - 1) -1(x - 1) = 0}$

$\implies\sf{(x - 1)(x - 1) = 0}$

$\implies\sf{x = 1, 1}$

Hence, the value of x is 1.