Question

How do you solve this equation $4{x^2} - 4x + 1 = 0$?

Answer 1

Hint: The given equation is a quadratic equation and cannot be solve using the method of factorization. So, we will use quadratic formula to solve this equation. Quadratic formula states that if $a{x^2} + bx + c = 0$ is a quadratic equation, so the root of this equation is given by the formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ where $a$ and $b$ are the coefficients of ${x^2}$ and $x$ respectively and $c$ is a constant.

Complete step by step answer:
The given equation is a quadratic equation. Quadratic equations are those equations that consist of at least one term which is squared and also called a second degree equation. The general form of a quadratic equation is $a{x^2} + bx + c = 0$ where $a, b$ and $c$ are numerical constants or coefficients, and $x$ is an unknown variable.
To solve the given equation $4{x^2} - 4x + 1 = 0$ we will use the quadratic formula. i.e.,
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Compare the given quadratic equation $4{x^2} - 4x + 1 = 0$ by the general form of quadratic equation $a{x^2} + bx + c = 0$.
 We get $a = 4,\,\,b = - 4$ and $c = 1$
Putting these values in the quadratic formula. We get,
$ \Rightarrow x = \dfrac{{ - ( - 4) \pm \sqrt {{{( - 4)}^2} - 4 \times 4 \times 1} }}{{2 \times 4}}$
Simplifying the above equation. We get,
$ \Rightarrow x = \dfrac{{4 \pm \sqrt {16 - 16} }}{8}$
$ \Rightarrow x = \dfrac{{4 \pm \sqrt 0 }}{8}$
$ \Rightarrow x = \dfrac{4}{8}$
$ \Rightarrow x = \dfrac{1}{2}$
Hence, the value of $x = \dfrac{1}{2}$ for the given quadratic equation.

Note:
 In the given problem we can’t use the method of factorization to find the roots of $x$. We need to keep in mind that the coefficient of $x$ should be broken in such a way that its sum gives us the value of the coefficient $x$ and its product gives the value of the coefficient of ${x^2}$.