Question

Factorize ${x^2} + 3x + 12 = 0$

Answer 1

Hint:Quadratic Formula: $a{x^2} + bx + c = 0$ Here $a,\;b,\;c$are numerical coefficients. So to solve $x$ we have:$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ So in order to solve the above given question using quadratic formula we have to find the values of $a,\;b,\;c$corresponding to the given question.

Then by substituting the values in the above equation we can find the values for $x$ and thereby factorize it.

Complete step by step solution:
Given
${x^2} + 3x + 12 = 0............................\left( i \right)$

Here we can’t use the grouping method since on observing the given expression we can’t find the two numbers whose sum is $3$and product is$12$. So we use the Quadratic formula to factorize the given expression.

 Now we need to compare (i) to the general formula and find the values of unknowns. Then we have to use the equation to find $x$by substituting all the values needed in it and by that way we can solve the equation${x^2} + 3x + 12 = 0$.

So on comparing (i) to the general formula$a{x^2} + bx + c = 0$, we get:
$a = 1,\;b = 3,\;c = 12.....................\left( {ii} \right)$

Now to solve for $x$we have $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}......................\left( {iii}
\right)$

Now substituting (ii) in (iii) we get:
\[
\;\;\;\;x = \dfrac{{ - \left( 3 \right) \pm \sqrt {{{\left( 3 \right)}^2} - 4\left( {1 \times 12} \right)} }}{{2
\times 1}} \\
\Rightarrow x = \dfrac{{ - 3 \pm \sqrt {\left( 9 \right) - 4\left( {12} \right)} }}{2} \\
\Rightarrow x = \dfrac{{ - 3 \pm \sqrt {\left( 9 \right) - 48} }}{2} \\
\Rightarrow x = \dfrac{{ - 3 \pm \sqrt { - 39} }}{2}...........................\left( {iv} \right) \\
\]

Now we know that negative roots can be represented by imaginary numbers such that:
$i = \sqrt { - 1} .............\left( v \right)$

So substituting (v) in (iv), we get:
\[ \Rightarrow x = \dfrac{{ - 3 \pm \sqrt {39} i}}{2}\]

Now we see that there are two possibilities for ‘x’ such that:
\[
\Rightarrow x = \dfrac{{ - 3 + \sqrt {39} i}}{2}\;\;{\text{and}}\;\; \Rightarrow x = \dfrac{{ - 3 - \sqrt
{39} i}}{2} \\
\Rightarrow x = \dfrac{{ - 3}}{2} + \dfrac{i}{2}\sqrt {39} \;\;{\text{and}}\;\; \Rightarrow x = \dfrac{{ -
3}}{2} - \dfrac{i}{2}\sqrt {39} \\
\]

Therefore on factoring we get:
$\left( {x + \dfrac{{3 - \sqrt {39} i}}{2}} \right)\left( {x + \dfrac{{3 + \sqrt {39} i}}{2}} \right) = 0$

So on factorization of ${x^2} + 3x + 12 = 0$we get $\left( {x + \dfrac{{3 - \sqrt {39} i}}{2}} \right)\left( {x
+ \dfrac{{3 + \sqrt {39} i}}{2}} \right) = 0.$

Note: Quadratic formula is mainly used in conditions where grouping method cannot be used or when the polynomials cannot be reduced into some general identity.

Here we can’t use a grouping method since the Quadratic formula method is an easier and direct method in comparison to other methods.

Also while using the Quadratic formula when $\sqrt {{b^2} - 4ac} $ is a negative root then the corresponding answer would be a complex number.