Question

How do you solve ${x^2} - 11x + 28 = 0$ using the quadratic formula?

Answer 1

Hint: First mention the general quadratic equation and then mention the formula of the roots given by $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ and substitute the values as per the given equation to get the answer.

Complete step by step answer:
We are given that we are required to solve ${x^2} - 11x + 28 = 0$ using the quadratic formula.
The general quadratic equation is given by $a{x^2} + bx + c = 0$, where a, b and c are the real numbers.
The roots of this general quadratic equation is given by the following formula:-
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Now, if we compare the given quadratic equation ${x^2} - 11x + 28 = 0$ to the general quadratic equation $a{x^2} + bx + c = 0$, we will then obtain the following:-
$ \Rightarrow $a = 1, b = - 11 and c = 28
Now, putting these values in the formula of quadratics, we will then obtain the following:-
$ \Rightarrow x = \dfrac{{ - ( - 11) \pm \sqrt {{{( - 11)}^2} - 4(1)(28)} }}{{2(1)}}$
Simplifying some calculations in the above equation, we will then obtain the following equation:-
$ \Rightarrow x = \dfrac{{11 \pm \sqrt {121 - 112} }}{2}$
Simplifying the calculations inside the square – root in the above equation, we will then obtain the following equation:-
$ \Rightarrow x = \dfrac{{11 \pm \sqrt 9 }}{2}$
Simplifying the calculations inside the square – root further in the above equation, we will then obtain the following equation:-
$ \Rightarrow x = \dfrac{{11 \pm 3}}{2}$
Therefore, the values of x can be 7 and 4.
Hence, the roots are 4 and 7.

Note: You can solve the above problem using the below method of splitting the middle term like the following:-
We are given that we are required to solve ${x^2} - 11x + 28 = 0$ using the quadratics formula.
We can write the given equation as:-
$ \Rightarrow {x^2} - 4x - 7x + 28 = 0$
Taking x common from first two terms, we will then obtain the following equation:-
$ \Rightarrow x(x - 4) - 7x + 28 = 0$
Taking – 7 common from first the last two terms, we will then obtain the following equation:-
$ \Rightarrow x(x - 4) - 7(x - 4) = 0$
Now taking (x – 4) common, we will then obtain the following:-
$ \Rightarrow (x - 4)(x - 7) = 0$
Thus the roots are 4 and 7.