Question

How do you find the roots of ${x^2} + x = 56$?

Answer 1

Hint: We will use the method of splitting of the middle term and then split our middle term according to the other terms and then find factors and thus solve it to find roots.

Complete step-by-step answer:
We are given that we are required to solve ${x^2} + x = 56$.
If we bring 56 from addition in the right hand side to subtraction in the left hand side, we get:
We can write this equation as ${x^2} + x - 56 = 0$.
Now, we can write the above equation as following:-
$ \Rightarrow {x^2} + 8x - 7x - 56 = 0$
Now, we can club two terms together to obtain the following:-
$ \Rightarrow \left( {{x^2} + 8x} \right) + \left( { - 7x - 56} \right) = 0$
Now, we will get x common from the first bracket and – 7 from the next bracket in the above expression to get the following expression:-
$ \Rightarrow x\left( {x + 8} \right) - 7\left( {x + 8} \right) = 0$
Now if we take x + 8 common from both the terms, we will then obtain the following equation:-
$ \Rightarrow \left( {x + 8} \right)\left( {x - 7} \right) = 0$
Hence, we get either x = - 8 or x = 7.

Hence, the roots are -8 and -7.

Note:
The students must note that after the step $\left( {x + 8} \right)\left( {x - 7} \right) = 0$, we directly got the roots due to a concept which can be stated as follows:-
If a.b = 0, then either a = 0 or b = 0.
We just replaced a by (x + 8) and b by (x – 7).
There is an alternate way to do the same, which can be done as following:-
If you have an equation $a{x^2} + bx + c = 0$, then the roots of the equation are given by:-
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$, where $D = {b^2} - 4ac$
If we compare this to the given equation, we have a = 1, b = 1 and c = -56
So, we get: $D = 1 - 4 \times ( - 56)$
If we solve it, we will get: D = 225
Now, let us find the roots:-
$ \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {225} }}{2}$
Solving the square – root, we will then obtain the following:-
$ \Rightarrow x = \dfrac{{ - 1 \pm 15}}{2}$
Thus, we have x = 7 or – 8.