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