Question

What are the roots of the equation $ {{x}^{2}}-5x+1=0 $ ?

Answer 1

Hint: We have been given a quadratic equation of $ x $ as $ {{x}^{2}}-5x+1=0 $ . We use the quadratic formula to solve the value of the $ x $ . we have the solution in the form of $ x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $ for general equation of $ a{{x}^{2}}+bx+c=0 $ . We put the values and find the solution.

Complete step by step solution:
We know for a general equation of quadratic $ a{{x}^{2}}+bx+c=0 $ , the value of the roots of $ x $ will be $ x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $ . This is the quadratic equation solving method. The root part $ \sqrt{{{b}^{2}}-4ac} $ of $ x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $ is called the discriminant of the equation.
In the given equation we have $ {{x}^{2}}-5x+1=0 $ . The values of a, b, c is $ 1,-5,1 $ respectively.
We put the values and get $ x $ as \[x=\dfrac{-\left( -5 \right)\pm \sqrt{{{\left( -5 \right)}^{2}}-4\times 1\times 1}}{2\times 1}=\dfrac{5\pm \sqrt{21}}{2}\]
The roots of the equation are irrational numbers. So, values of x are $ x=\dfrac{5\pm \sqrt{21}}{2} $ .
The discriminant value being not a perfect square, we get the irrational numbers as root values.
In this case the value of $ D=\sqrt{{{b}^{2}}-4ac} $ is non-square. $ {{b}^{2}}-4ac={{\left( -5 \right)}^{2}}-4\times 1\times 1=21 $ .
This is a square root value of 21. That’s why the roots are irrational.
So, the correct answer is “ $ x=\dfrac{5\pm \sqrt{21}}{2} $ .”.

Note: We have been given the equation $ {{x}^{2}}-5x+1=0 $ . We form the square part in $ {{x}^{2}}-5x+1 $ .
The square form of subtraction of two numbers be $ {{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} $ .
We have $ {{x}^{2}}-5x+1={{x}^{2}}-2\times x\times \dfrac{5}{2}+{{\left( \dfrac{5}{2} \right)}^{2}}+1-{{\left( \dfrac{5}{2} \right)}^{2}} $ .
Forming the square, we get $ {{x}^{2}}-5x+1={{\left( x-\dfrac{5}{2} \right)}^{2}}-{{\left( \dfrac{\sqrt{21}}{2} \right)}^{2}} $ .
We get $ {{\left( x-\dfrac{5}{2} \right)}^{2}}-{{\left( \dfrac{\sqrt{21}}{2} \right)}^{2}}=0 $ . Taking solution, we get
 $
  {{\left( x-\dfrac{5}{2} \right)}^{2}}-{{\left( \dfrac{\sqrt{21}}{2} \right)}^{2}}=0 \\
 \Rightarrow {{\left( x-\dfrac{5}{2} \right)}^{2}}={{\left( \dfrac{\sqrt{21}}{2} \right)}^{2}} \\
 \Rightarrow \left( x-\dfrac{5}{2} \right)=\pm \dfrac{\sqrt{21}}{2} \\
 \Rightarrow x=\dfrac{5\pm \sqrt{21}}{2} \;
 $ .
Thus, the solution of the equation $ {{x}^{2}}-5x+1=0 $ is $ x=\dfrac{5\pm \sqrt{21}}{2} $ .