Question

How do you solve the equation $x^2+4x=21$?

Answer 1

Hint: Factorizing reduces the higher degree equation into its linear equation. In the above given question, we need to reduce the quadratic equation into its simplest form in such a way that addition of products of the factors of first and last term should be equal to the middle term i.e. $4x$.

Complete step by step answer:
The equation $a{x}^2+bx+c$ is a general way of writing quadratic equations where a, b and c are numbers.
In the above expression, can be written as
$\Rightarrow x^2+4x-21=0$
a=1, b=4, c=-21
First step is by multiplying the coefficient of $x^2$ and the constant term -21, we get $-21x^2$.
After this, factors of $-21x^2$should be calculated in such a way that their addition should be equal to 4x.
Factors of 21 can be 7 and -3.
Since, we consider negative sign of 21 one of the factors should be negative so that multiplication result is negative and by adding these two factors we should get positive $$14x$$.
Therefore, we can add in the following way $7x^{}$+ $(-3x)$=$4x$.
So, further we write the equation by equating it with zero and splitting the middle term according to the factors.
$$\begin{gathered} \Rightarrow x^2+4x-21=0 \\ \Rightarrow x^2+7x-3x-21=0 \end{gathered}$$
Now, by grouping the first two and last two terms we get common factors.
$$\begin{gathered} \Rightarrow x(x+7)-3(x+7)=0 \\ \Rightarrow (x-3)(x+7)=0 \end{gathered}$$
Taking x common from the first group and -3 common from the second we get the above equation.
Therefore, by solving the above quadratic equation we get factors 3 and - 7.

Note: In quadratic equation, an alternative way of finding the factors is by using a formula which is given below:
$\Rightarrow x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$
By substituting the values of a=1, b=4 and c=-21 we get the factors of x.
$\Rightarrow x={\displaystyle \frac{-(4)\pm \sqrt{{(4)}^2-4(1)(-21)}}{2(1)}}$
So, the values are $$x=3$$ or $$x=-7$$.