Question

Find the roots of the quadratic equation $ {{x}^{2}}-2x+1=0 $ .

Answer 1

Hint: Here, in this given question, we can use the quadratic formula for the roots of the general quadratic equation $ a{{x}^{2}}+bx+c=0 $ , which is $ x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $ in order to find the roots of the given equation. Just we have to substitute the values of a, b and c with $ 1 $ , $ -2 $ and $ 1 $ respectively and we will obtain the required roots of the given quadratic equation $ {{x}^{2}}-2x+1=0 $ .

Complete step-by-step answer:
In this given question, we are asked to find out the roots of the given quadratic equation
 $ {{x}^{2}}-2x+1=0 $ ………….. (1.1)
A quadratic equation is an equation with the highest power of x as 2 or whose degree is 2.
Now, we are going to use the quadratic formula for the roots of the general quadratic equation $ a{{x}^{2}}+bx+c=0 $ , which is $ x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $ ………………(1.2), in order to find the roots of the given equation.
In equation 1.1, the values of a, b and c are $ 1 $ , $ -2 $ and $ 1 $ respectively.
So, putting these values in equation 1.2, we get,
 $ x=\dfrac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\times 1\times 1}}{2\times 1}...........(1.3) $
Now, simplifying equation 1.3, we get,
 $ \begin{align}
  & x=\dfrac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\times 1\times 1}}{2\times 1} \\
 & =\dfrac{2\pm \sqrt{4-4}}{2} \\
 & =\dfrac{2\pm \sqrt{0}}{2} \\
 & =\dfrac{2}{2}=1...............(1.4) \\
\end{align} $
Hence, we have obtained the value of x as 1.
This has happened because $ {{x}^{2}}-2x+1=0 $ is the square of $ \left( x-1 \right) $ . So, this has only one root as 1.
Therefore, the root of the equation $ {{x}^{2}}-2x+1=0 $ is 1,1.

Note: Although we have used the quadratic formula to find the roots of the given equation, we must note that there are also methods like factorization, completing the square and graphical method to solve this sort of question. We can use any of the methods; however the answer will remain the same.