Question

Solve $x=\dfrac{1}{2-\dfrac{1}{2-\dfrac{1}{2-x}}}.$ \[\]

Answer 1

Hint: We begin from the bottom of the right hand side of the equation using the formula $a-\dfrac{b}{c}=\dfrac{ac-b}{c}$ and then we use the reciprocal as $\dfrac{1}{\dfrac{a}{b}}=\dfrac{b}{a}$. We simplify and cross-multiply to get a quadratic equation in $x$ which we solve by completing the square method. \[\]

Complete step-by-step answer:
We know from addition and subtraction of fractions that for some real numbers $a,b,c$ where none of them are zero, we have
\[a-\dfrac{b}{c}=\dfrac{ac-b}{c},\dfrac{1}{\dfrac{a}{b}}=\dfrac{b}{a}\]
We know that when we are asked to solve and given an equation involving unknown variables then we have to find the value of unknown variables. Here in the question the unknown variable is $x$ and the expression is
\[x=\dfrac{1}{2-\dfrac{1}{2-\dfrac{1}{2-x}}}......\left( 1 \right)\]

Let us simplify from the bottom of the right hand side of the equation using the formula $a-\dfrac{b}{c}=\dfrac{ac-b}{c}$ for $a=2,b=1,c=2-x.$We have,
\[\Rightarrow x=\dfrac{1}{2-\dfrac{1}{\dfrac{2\left( 2-x \right)-1}{2-x}}}=\dfrac{1}{2-\dfrac{1}{\dfrac{3-x}{2-x}}}\]

We use the formula $\dfrac{1}{\dfrac{a}{b}}=\dfrac{b}{a}$ in the above step for $a=3-x,b=2-x$ and have,
\[\Rightarrow x==\dfrac{1}{2-\dfrac{2-x}{3-x}}\]
Let us simplify the denominator at the right hand side of the equation using the formula $a-\dfrac{b}{c}=\dfrac{ac-b}{c}$ for $a=2,b=2-x,c=3-x.$ We have
\[\Rightarrow x==\dfrac{1}{\dfrac{2\left( 3-x \right)-\left( 2-x \right)}{3-x}}=\dfrac{1}{\dfrac{4-3x}{3-x}}\]
We use the formula $\dfrac{1}{\dfrac{a}{b}}=\dfrac{b}{a}$ in the above step for $a=4-3x,b=2-x$ and have,
\[\Rightarrow x=\dfrac{3-2x}{4-3x}\]
Let us cross multiply across the equation and have
\[\begin{align}
  & \Rightarrow x\left( 4-3x \right)=1\times \left( 3-2x \right) \\
 & \Rightarrow 4x-3{{x}^{2}}=3-2x \\
 & \Rightarrow 3{{x}^{2}}-6x+3=0 \\
\end{align}\]
Let us divide both side of the equation by 3 and have
\[\begin{align}
  & \Rightarrow {{x}^{2}}-2x+1=0...\left( 2 \right) \\
 & \Rightarrow {{\left( x \right)}^{2}}-2\times x\times 1+{{\left( 1 \right)}^{2}}=0 \\
\end{align}\]
The equation (2) is a quadratic equation because the highest degree on $x$ is 2 which we are going to solve by completing the square method. Let us use the algebraic identity ${{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}$ for $a=x,b=1$ and have a complete square as
\[\begin{align}
  & {{\left( x-1 \right)}^{2}}=0 \\
 & \Rightarrow \left( x-1 \right)\left( x-1 \right)=0 \\
 & \Rightarrow x-1=0\text{ or }x-1=0 \\
 & \Rightarrow x=1,1 \\
\end{align}\]
So we have obtained two values for $x$ which are equal.

Note: We can alternatively solve the quadratic equation $a{{x}^{2}}+bx+c=;a\ne 0,b,c\in R$by splitting the middle term $b$or using the quadratic formula$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . The value of $x$ obtained by solving an equation are called zeroes or roots. The condition for two equal just like obtained in the problem is ${{b}^{2}}-4ac=0.$