Question

How do you solve ${{x}^{2}}+3x+2=0$?

Answer 1

Hint: For solving the above problem, we have to use the middle term factorization. After which we will get two values of x for our problem. In the middle term factorization, we write the middle term in such a way of addition or subtraction that we can take common and make pairs to find the solution.

Complete step by step solution:
The above problem is ${{x}^{2}}+3x+2=0$
We have to first solve this problem with factorising means we have to split the middle term
 $\Rightarrow {{x}^{2}}+3x+2=0$
 $\Rightarrow {{x}^{2}}+2x+x+2=0$
After taking common,
 $\Rightarrow x(x+2)+1(x+2)=0$
 $\Rightarrow (x+1)(x+2)=0$
Equating both the factors with zero
When x + 1 = 0
 $\Rightarrow x+1=0$
 $\therefore x=-1$
When x + 2 = 0
 $\Rightarrow x+2=0$
 $\therefore x=-2$

The values of x are either -1 or -2

Additional Information:
The above question was in the standard form of a quadratic equation that is $a{{x}^{2}}+bx+c=0$ . Remember that if the coefficients of x have any negative sign then we have to consider that sign as well.

Note:
There is an alternative way of solving this problem. Let us see that as well
We will use Shreedhara Acharya's formula that is $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .
Our problem is in standard form $a{{x}^{2}}+bx+c=0$ , where a = 1, b = 3 and c = 2.
By putting the values in the above formula we get
$\Rightarrow x=\dfrac{-3\pm \sqrt{{{3}^{2}}-4.1.2}}{2.1}$
When we take the plus sign, we get,
$\Rightarrow x=\dfrac{-3+\sqrt{9-8}}{2}$
$\therefore x=-1$
When we take the minus sign, we get,
$\Rightarrow x=\dfrac{-3-\sqrt{9-8}}{2}$
$\Rightarrow x=-2$
The values of x are either -1 or -2.