Question

How to factorise ${x^2} - 2x + 4 = 0$?

Answer 1

Hint:Any equation in x with power greater than one may be factored. After factorization, the equation can be written as the product of two equations of x with smaller power as compared with the original equation. It helps to make the calculation easier as the equations are smaller and can be easily noticed.

Complete step by step solution:
According to the question we have to factorise ${x^2} - 2x + 4 = 0$,
So, at first we have to find the solution of this equation ${x^2} - 2x + 4 = 0$, so we use the formula-
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ , where a, b and c are as given in the equation $a{x^2} + bx + c = 0$
When we compare $a{x^2} + bx + c = 0$ with ${x^2} - 2x + 4 = 0$ we will find that
$\begin{array}{l}
a = 1\\
b = - 2\\
c = 4
\end{array}$
Now, put these values in the formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
So, $x = \dfrac{{ - ( - 2) \pm \sqrt {4 - 4(1)(4)} }}{{2(1)}}$
$x = \dfrac{{2 \pm \sqrt { - 12} }}{2}$
As the number is negative inside the square root is negative, so there will be no real root of this equation.
So this equation can’t be factorized.
We can also find if any equation can be factored or not by just observing the value of $D$
Where$D = \sqrt {{b^2} - 4ac} $, where a, b and c are as given in the equation $a{x^2} + bx + c = 0$
When $D > 0 \Rightarrow $ equation can be factored and have two solutions
When $D < 0 \Rightarrow $ equation can’t be factored
When $D = 0 \Rightarrow $ equation can be factored and have only one solution
So, in this case, $D < 0$ so can’t be factored.

Note: We can use these formulas to find the roots very easily. We can factorize even in the form of imaginary numbers and get these as the roots of the equation. But these solutions are not real and are not considered in this question.