Question

How do you solve ${{x}^{2}}-4x+1=0$ using the quadratic formula?

Answer 1

Hint: We have been given a quadratic equation of x as ${{x}^{2}}-4x+1=0$. 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 answer:
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 ${{x}^{2}}-4x+1=0$. The values of a, b, c are $1,-4,1$ respectively.
We put the values and get x as \[x=\dfrac{-\left( -4 \right)\pm \sqrt{{{\left( -4 \right)}^{2}}-4\times 1\times 1}}{2\times 1}=\dfrac{4\pm \sqrt{12}}{2}=2\pm \sqrt{3}\].
The roots of the equation are irrational numbers.
The discriminant value being non-square, we get the irrational numbers a root value.
In this case the value of $D=\sqrt{{{b}^{2}}-4ac}$ is non-square. ${{b}^{2}}-4ac={{\left( -4 \right)}^{2}}-4\times 1\times 1=12$.
This is a non-square value. That’s why the roots are irrational.

Note: We find the value of x for which the function $f\left( x \right)={{x}^{2}}-4x+1$. We can see $f\left( 2+\sqrt{3} \right)={{\left( 2+\sqrt{3} \right)}^{2}}-4\left( 2+\sqrt{3} \right)+1=4+3+4\sqrt{3}-8-4\sqrt{3}+1=0$. So, the root of the $f\left( x \right)={{x}^{2}}-4x+1$ will be the $2+\sqrt{3}$. This means for $x=a$, if $f\left( a \right)=0$ then $\left( x-a \right)$ is a root of $f\left( x \right)$. We can also do the same process for $2-\sqrt{3}$.
We can also solve using the square form.
We have ${{x}^{2}}-4x+1={{\left( x-2 \right)}^{2}}-3$.
We get ${{\left( x-2 \right)}^{2}}-3=0$. Taking solution, we get
$\begin{align}
  & {{\left( x-2 \right)}^{2}}-3=0 \\
 & \Rightarrow {{\left( x-2 \right)}^{2}}=3 \\
 & \Rightarrow \left( x-2 \right)=\pm \sqrt{3} \\
 & \Rightarrow x=2\pm \sqrt{3} \\
\end{align}$.
Thus, verified the solution of the equation ${{x}^{2}}-4x+1=0$ is $x=2\pm \sqrt{3}$.