Question

How do you solve $4{x^2} - 5x + 1 = 0$?

Answer 1

Hint: This problem deals with solving a quadratic equation. Here, given a quadratic equation expression, we have to simplify the expression and make it into a standard form of quadratic equation. If the quadratic equation is in the form of $a{x^2} + bx + c = 0$, then we know that the roots of this quadratic equation are given by :
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step-by-step solution:
The given equation is $4{x^2} - 5x + 1 = 0$, consider it as given below:
$ \Rightarrow 4{x^2} - 5x + 1 = 0$
Now the above equation is in the standard form of a quadratic equation, which is $a{x^2} + bx + c = 0$.
Here comparing the equation $4{x^2} - 5x + 1 = 0$ with the standard form $a{x^2} + bx + c = 0$ and compare the coefficients $a,b$ and $c$:
$ \Rightarrow a = 4,b = - 5$ and $c = 1$
Now applying the formula to find the value of the roots of $x$, as given below:
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Substituting the values of $a,b$ and $c$ in the above formula:
$ \Rightarrow x = \dfrac{{ - \left( { - 5} \right) \pm \sqrt {{{\left( { - 5} \right)}^2} - 4\left( 4 \right)\left( 1 \right)} }}{{2(4)}}$
$ \Rightarrow x = \dfrac{{5 \pm \sqrt {25 - 16} }}{8}$
Simplifying the above expression, as given below:
$ \Rightarrow x = \dfrac{{5 \pm \sqrt 9 }}{8}$
We know that the square root of 9 is 3, $\sqrt 9 = 3$
$ \Rightarrow x = \dfrac{{5 \pm 3}}{8}$
Now considering the two cases, with plus and minus, as shown:
$ \Rightarrow x = \dfrac{{5 + 3}}{8};x = \dfrac{{5 - 3}}{8}$
$ \Rightarrow x = \dfrac{8}{8};x = \dfrac{2}{8}$
Hence the value of the roots are equal to :
$\therefore x = 1;x = \dfrac{1}{4}$
The solution of the equation $4{x^2} - 5x + 1 = 0$ are $1$ and $\dfrac{1}{4}$.

Note: Please note that this problem can also be done either by the method of completing the square or just factoring and solving the quadratic equation. To solve $a{x^2} + bx + c = 0$ by completing the square: transform the equation so that the constant term,$c$ is alone on the right side. But here we are adding and subtracting some terms in order to factor.