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- #1

- Feb 14, 2012

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

- Thread starter anemone
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- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,937

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

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- #2

- Jan 26, 2012

- 4,202

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

The quadratic formula yields

$$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

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

- Jan 17, 2013

- 1,667

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

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

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

Hence :

\(\displaystyle x=-1\) or

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