We are given the inequality $\frac{m^2 - 20}{-(2m+5)} - \frac{m^2 + m - 12}{2m+5} < 2$ and the condition that $m \neq -\frac{5}{2}$. We need to simplify the inequality.

AlgebraInequalitiesQuadratic EquationsRational ExpressionsInterval Notation
2025/5/20

1. Problem Description

We are given the inequality m220(2m+5)m2+m122m+5<2\frac{m^2 - 20}{-(2m+5)} - \frac{m^2 + m - 12}{2m+5} < 2 and the condition that m52m \neq -\frac{5}{2}. We need to simplify the inequality.

2. Solution Steps

First, combine the fractions on the left side of the inequality:
m220(2m+5)m2+m122m+5<2\frac{m^2 - 20}{-(2m+5)} - \frac{m^2 + m - 12}{2m+5} < 2
(m220)2m+5m2+m122m+5<2\frac{-(m^2 - 20)}{2m+5} - \frac{m^2 + m - 12}{2m+5} < 2
m2+20m2m+122m+5<2\frac{-m^2 + 20 - m^2 - m + 12}{2m+5} < 2
2m2m+322m+5<2\frac{-2m^2 - m + 32}{2m+5} < 2
Now, subtract 2 from both sides:
2m2m+322m+52<0\frac{-2m^2 - m + 32}{2m+5} - 2 < 0
2m2m+322(2m+5)2m+5<0\frac{-2m^2 - m + 32 - 2(2m+5)}{2m+5} < 0
2m2m+324m102m+5<0\frac{-2m^2 - m + 32 - 4m - 10}{2m+5} < 0
2m25m+222m+5<0\frac{-2m^2 - 5m + 22}{2m+5} < 0
(2m2+5m22)2m+5<0\frac{-(2m^2 + 5m - 22)}{2m+5} < 0
2m2+5m222m+5>0\frac{2m^2 + 5m - 22}{2m+5} > 0
Let's find the roots of the quadratic equation 2m2+5m22=02m^2 + 5m - 22 = 0. Using the quadratic formula,
m=b±b24ac2am = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
m=5±524(2)(22)2(2)m = \frac{-5 \pm \sqrt{5^2 - 4(2)(-22)}}{2(2)}
m=5±25+1764m = \frac{-5 \pm \sqrt{25 + 176}}{4}
m=5±2014m = \frac{-5 \pm \sqrt{201}}{4}
So the roots are m1=520144.79m_1 = \frac{-5 - \sqrt{201}}{4} \approx -4.79 and m2=5+20142.29m_2 = \frac{-5 + \sqrt{201}}{4} \approx 2.29.
Also, 2m+5=02m+5 = 0 when m=52=2.5m = -\frac{5}{2} = -2.5.
Now, we analyze the sign of 2m2+5m222m+5\frac{2m^2 + 5m - 22}{2m+5}. We have three critical points: m14.79m_1 \approx -4.79, m=2.5m = -2.5 and m22.29m_2 \approx 2.29.
We test the sign of the expression in the intervals (,4.79)(-\infty, -4.79), (4.79,2.5)(-4.79, -2.5), (2.5,2.29)(-2.5, 2.29), and (2.29,)(2.29, \infty).

1. $m < -4.79$, let $m = -5$.

2(5)2+5(5)222(5)+5=50252210+5=35<0\frac{2(-5)^2 + 5(-5) - 22}{2(-5)+5} = \frac{50 - 25 - 22}{-10+5} = \frac{3}{-5} < 0.

2. $-4.79 < m < -2.5$, let $m = -3$.

2(3)2+5(3)222(3)+5=1815226+5=191>0\frac{2(-3)^2 + 5(-3) - 22}{2(-3)+5} = \frac{18 - 15 - 22}{-6+5} = \frac{-19}{-1} > 0.

3. $-2.5 < m < 2.29$, let $m = 0$.

2(0)2+5(0)222(0)+5=225<0\frac{2(0)^2 + 5(0) - 22}{2(0)+5} = \frac{-22}{5} < 0.

4. $m > 2.29$, let $m = 3$.

2(3)2+5(3)222(3)+5=18+15226+5=1111>0\frac{2(3)^2 + 5(3) - 22}{2(3)+5} = \frac{18 + 15 - 22}{6+5} = \frac{11}{11} > 0.
Since we want 2m2+5m222m+5>0\frac{2m^2 + 5m - 22}{2m+5} > 0, the solution is (4.79,2.5)(2.29,)(-4.79, -2.5) \cup (2.29, \infty), or (52014,52)(5+2014,)(\frac{-5 - \sqrt{201}}{4}, -\frac{5}{2}) \cup (\frac{-5 + \sqrt{201}}{4}, \infty).

3. Final Answer

The solution to the inequality m220(2m+5)m2+m122m+5<2\frac{m^2 - 20}{-(2m+5)} - \frac{m^2 + m - 12}{2m+5} < 2 is m(52014,52)(5+2014,)m \in (\frac{-5 - \sqrt{201}}{4}, -\frac{5}{2}) \cup (\frac{-5 + \sqrt{201}}{4}, \infty).
Final Answer: m(52014,52)(5+2014,)m \in (\frac{-5-\sqrt{201}}{4}, -\frac{5}{2}) \cup (\frac{-5+\sqrt{201}}{4}, \infty)

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