Question

How do you solve\[2{x^2} - 4x + 1 = 0\]?

Answer 1

Hint: Use method of determinant to solve for the value of x from the given quadratic equation. Compare the quadratic equation with general quadratic equation and substitute values in the formula of finding roots of the equation.
* For a general quadratic equation \[a{x^2} + bx + c = 0\], roots are given by formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]

Complete step-by-step answer:
We are given the quadratic equation \[2{x^2} - 4x + 1 = 0\] … (1)
We know that general quadratic equation is \[a{x^2} + bx + c = 0\]
On comparing with general quadratic equation \[a{x^2} + bx + c = 0\], we get \[a = 2,b = - 4,c = 1\]
Substitute the values of a, b and c in the formula of finding roots of the equation.
\[ \Rightarrow x = \dfrac{{ - ( - 4) \pm \sqrt {{{( - 4)}^2} - 4 \times 2 \times 1} }}{{2 \times 2}}\]
Square the values inside the square root in numerator of the fraction
\[ \Rightarrow x = \dfrac{{4 \pm \sqrt {16 - 8} }}{4}\]
Calculate the value under square root in the fraction
\[ \Rightarrow x = \dfrac{{4 \pm \sqrt 8 }}{4}\]
We know that \[8 = {2^2} \times 2\]
Substitute this value of 8 under the square root in the numerator
\[ \Rightarrow x = \dfrac{{4 \pm \sqrt {{2^2} \times 2} }}{4}\]
Cancel square root by square power in the numerator
\[ \Rightarrow x = \dfrac{{4 \pm 2\sqrt 2 }}{4}\]
Take 2 common from both terms in the numerator
\[ \Rightarrow x = \dfrac{{2(2 \pm \sqrt 2 )}}{4}\]
Cancel same factors from numerator and denominator i.e. 2
\[ \Rightarrow x = \dfrac{{2 \pm \sqrt 2 }}{2}\]
So, \[x = \dfrac{{2 + \sqrt 2 }}{2}\]and \[x = \dfrac{{2 - \sqrt 2 }}{2}\]

\[\therefore \]Solution of the equation \[2{x^2} - 4x + 1 = 0\] is \[x = \dfrac{{2 + \sqrt 2 }}{2}\]and \[x = \dfrac{{2 - \sqrt 2 }}{2}\].

Note:
Many students make the mistake of solving for the roots or values of x for this question using factorization method which is wrong. They open the coefficient of x as \[ - 2x - 2x\] but this is wrong as it will give us sum of \[ - 4x\] but it will not give us the product of terms to be coefficient of \[{x^2}\]. Keep in mind the coefficient of x should be broken in such a manner that its sum gives the value of coefficient of x and its product gives the value of coefficient of \[{x^2}\].