Welcome to our community

anemone

MHB POTW Director
Staff member
Solve $$\displaystyle (m-2)x^2-(m+3)x-2m-1=0$$.

Ackbach

Indicium Physicus
Staff member
Solve $$\displaystyle (m-2)x^2-(m+3)x-2m-1=0$$.
Before beginning, we note that $m \not=2$, or else the equation is not quadratic but linear. So, we assume $m \not=2$.
$$x= \frac{m+3 \pm \sqrt{(m+3)^{2}-4(m-2)(-2m-1)}}{2(m-2)} = \frac{m+3 \pm \sqrt{m^{2}+6m+9-4(-2m^{2}+3m+2)}}{2(m-2)}$$
$$= \frac{m+3 \pm \sqrt{9m^{2}-6m+1}}{2(m-2)} = \frac{m+3 \pm \sqrt{(3m-1)^{2}}}{2(m-2)}= \frac{m+3 \pm |3m-1|}{2(m-2)}.$$
These two solutions will not change, actually, depending on whether $m<1/3$ or $m \ge 1/3$, since we're multiplying the absolute value by $\pm$. Hence, we have the solutions

$$x= \left \{-1,\;\frac{2m+1}{m-2} \right \}.$$

ZaidAlyafey

Well-known member
MHB Math Helper
By some rearrangements :

$$\displaystyle m(x^2-x-2)-(2x^2+3x+1)=0$$

$$\displaystyle m(x-2)(x+1)-(2x+1)(x+1)=0$$

$$\displaystyle (x+1) \left(m(x-2)-(2x+1)\right)=0$$

Hence :

$$\displaystyle x=-1$$ or

$$\displaystyle x=\frac{2m+1}{m-2}$$