Question

Solve the quadratic equation \[{{x}^{2}}-4x+2=0\] by the formula method.

Answer 1

Hint: The roots of a quadratic equation of the form \[a{{x}^{2}}+bx+c=0\] is given by \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] . Now, compare the quadratic equation \[{{x}^{2}}-4x+2=0\] and the standard form of the quadratic equation which is \[a{{x}^{2}}+bx+c=0\] . Put the values of a,b, and c in the formula \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] and get the values of x.

Complete step-by-step answer:

According to the question, it is given that our quadratic equation is \[{{x}^{2}}-4x+2=0\] ……………………(1)
We have to find the roots of the quadratic equation \[{{x}^{2}}-4x+2=0\] using the formula method.
We know the standard form of the quadratic equation which is \[a{{x}^{2}}+bx+c=0\] ………………..(2)
From equation (1), we have our quadratic equation.
Now, we have to compare equation (1) and equation (2) so that we can relate the quadratic equation \[{{x}^{2}}-4x+2=0\] and the standard form of the quadratic equation \[a{{x}^{2}}+bx+c=0\] .
Comparing \[{{x}^{2}}-4x+2=0\] and the standard quadratic equation \[a{{x}^{2}}+bx+c=0\] , we get
\[a=1\] ……………..(3)
\[b=-4\] ………………….(4)
\[c=2\] ……………………(5)
We know the formula to get the roots of the standard form of the quadratic equation,
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] …………………(6)
From equation (3), equation (4) and equation (5), we have to put the values of a, b, and c in equation (6).
Now, putting the values of a, b, and c in equation (6), we get
\[\begin{align}
  & x=\dfrac{-(-4)\pm \sqrt{{{(-4)}^{2}}-4.1.2}}{2.1} \\
 & \Rightarrow x=\dfrac{4\pm \sqrt{16-8}}{2} \\
 & \Rightarrow x=\dfrac{4\pm \sqrt{8}}{2} \\
 & \Rightarrow x=2\pm \sqrt{2} \\
\end{align}\]
Hence, the roots of the quadratic equation \[({{x}^{2}}-4x+2=0)\] are \[(2+\sqrt{2})\] and \[(2-\sqrt{2})\] .

Note: We can also solve this question using the perfect square method. We know the formula,
\[{{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\] ………………………(1)
Our given quadratic equation is \[({{x}^{2}}-4x+2=0)\] ……………..(2)
Making the quadratic equation \[({{x}^{2}}-4x+2=0)\] as a perfect square,
\[{{x}^{2}}-4x+2=0\]
\[\Rightarrow {{x}^{2}}-2.2.x+2+2=2\]
\[\Rightarrow {{x}^{2}}-2.2.x+{{2}^{2}}=2\] …………………..(3)
 Comparing equation (2) and equation (3), we get
\[a=x\] ………………(4)
\[b=2\] …………….(5)
Now, using the formula \[{{(a-b)}^{2}}=({{a}^{2}}+{{b}^{2}}-2ab)\] , transforming equation (3), we get
\[\Rightarrow {{x}^{2}}-2.2.x+{{2}^{2}}=2\]
\[\begin{align}
  & \Rightarrow {{(x-2)}^{2}}=2 \\
 & \Rightarrow (x-2)=\pm \sqrt{2} \\
 & \Rightarrow x=2\pm \sqrt{2} \\
\end{align}\]
Hence, the roots of the quadratic equation \[({{x}^{2}}-4x+2=0)\] are \[(2+2\sqrt{2})\] and \[(2-2\sqrt{2})\] . This is one way we can verify our result. Since we have been asked to use the formula method, we must stick to the quadratic formula to find the roots.