Question

How do you factorize and solve, ${{m}^{2}}=-6m-7$ ?

Answer 1

To completely factorize and solve a quadratic expression, we will first of all write the quadratic in its standard form and then try to break the middle term into two individual terms. If this doesn’t work, then we shall use the formula for finding the roots of a quadratic and then write it in its factored form. We shall proceed like this to get the required answer to our problem.

Complete step-by-step solution:
We have been given to solve a quadratic equation which goes like:
$\Rightarrow {{m}^{2}}=-6m-7$
First of all, we will write this quadratic equation in its standard form. And, the standard form of a quadratic equation is given by:
$\Rightarrow a{{x}^{2}}+bx+c=0$
Writing our quadratic equation in standard form, we get:
$\Rightarrow {{m}^{2}}+6m+7=0$
Now, on comparing our standard equation with the generalized form of standard quadratic equation, we get:
$\begin{aligned} & \Rightarrow a=1 \\\ & \Rightarrow b=6 \\\ & \Rightarrow c=7 \\\ \end{aligned}$
Here, we can see that the middle term of our quadratic equation cannot be split into individual terms to get a common factor. So, we will use the formula for finding the roots of the quadratic.
For any quadratic equation in its generalized form, the roots of the quadratic are given by:
\Rightarrow \left\\{ \alpha ,\beta \right\\}=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}
Where, ‘$\alpha$’ and ‘$\beta$’ are the two roots of our quadratic.
Applying this formula on our quadratic equation, we get the roots of our quadratic equation as:
$\begin{aligned} & \Rightarrow \alpha =\dfrac{-6+\sqrt{{{6}^{2}}-4\times 1\times 7}}{2\times 1} \\\ & \Rightarrow \alpha =\dfrac{-6+\sqrt{8}}{2} \\\ & \Rightarrow \alpha =\dfrac{-6+2\sqrt{2}}{2} \\\ & \Rightarrow \alpha =-3+\sqrt{2} \\\ \end{aligned}$
Now, since this is an irrational root, the irrational part of the second root will be a conjugate of the first root. Thus, we have:
$\Rightarrow \beta =-3-\sqrt{2}$
Thus, we have solved our quadratic equation and found the roots or solutions of our quadratic equation as, $\left( -3+\sqrt{2} \right)\text{ and }\left( -3-\sqrt{2} \right)$ .
Thus, our quadratic equation could be written in its factor form as follows:
$\begin{aligned} & \Rightarrow \left( m-\alpha \right)\left( m-\beta \right)=0 \\\ & \Rightarrow \left[ m-\left( -3+\sqrt{2} \right) \right]\left[ m-\left( -3-\sqrt{2} \right) \right]=0 \\\ & \Rightarrow \left( m+3-\sqrt{2} \right)\left( m+3+\sqrt{2} \right)=0 \\\ \end{aligned}$
Thus, the factored form of our quadratic equation, ${{m}^{2}}=-6m-7$, comes out to be $\Rightarrow \left( m+3-\sqrt{2} \right)\left( m+3+\sqrt{2} \right)=0$
Hence, we have factorized our quadratic equation by first find its root and these roots are $\left( -3+\sqrt{2} \right)\text{ and }\left( -3-\sqrt{2} \right)$.

Note: The factorization method of finding roots by splitting the central term of a quadratic is the easiest method to solve a quadratic. But, it has a catch. When, the roots are irrational, we cannot use it to solve our quadratic equation and one in such case one should rather use the formula for finding roots of the quadratic equation as it has no exceptions to its application.