How would one factorise $m^2 + (2AB)m + B^2 =0$

Question

How would one factorise $m^2 + (2AB)m + B^2 =0$, to go onto solve a second order differential equation

Answer 1

Hint: $$m^2 + 2ABm + B^2 = \left(m^2 + 2ABm + A^2B^2\right) + B^2 - A^2B^2$$ $$=(m+AB)^2 + B^2 - A^2B^2 = 0$$

Answer 2

We have

$$m^2 + (2AB)m + B^2 =0\implies m=\frac{-2AB\pm \sqrt{4A^2B^2-4B^2}}{2}=-AB\pm B\sqrt{A^2-1}$$

and therefore

$$m^2 + (2AB)m + B^2=[m-(-AB+B\sqrt{A^2-1})]\cdot[m-(-AB-B\sqrt{A^2-1})]$$

Answer 3

Let $$ r_1 = -AB + B\sqrt{A^2 - 1}\\ r_2 = -AB - B\sqrt{A^2 - 1} $$ Then your polynomial is $$ (m-r_1)(m-r_2). $$

Answer 4

$$m^2+2ABm+B^2=(m+AB)^2-A^2B^2+B^2 \\=\left(m+AB-\sqrt{A^2B^2-B^2}\right)\left(m+AB+\sqrt{A^2B^2-B^2}\right)$$

provided $|A|\ge1$. Otherwise you need complex numbers.

Answer 5

If $m$ is your variable, then $$m^2 + (2AB)m + B^2\tag{1}$$ is a 2nd degree polynomial in $m$, and you can factorize it by finding the roots of $(1)$. Using the quadratic formula, the roots are: $$m=\frac{-2AB\pm\sqrt{(2AB)^2-4B^2}}{2}=-AB\pm B\sqrt{A^2-B^2}$$ So $(1)$ may be written as $$\left(m+AB-B\sqrt{A^2-B^2}\right)\left(m+AB+B\sqrt{A^2-B^2}\right)$$

Answer 6

Hint: Note that

$$m^2+(2AB)m+B^2 = (m-m_1)(m-m_2)$$

if $m_1$ and $m_2$ are the roots of the equation. Also, recall

$$ax^2+bx+c = 0 \iff x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

which you can apply to $m^2+(2AB)m+B^2 = 0$ to solve for $m$, getting you $m_1$ and $m_2$.