Question 30

Both roots of the quadratic equation $(m + 2)x^2 - (2m - 1)x + m + 1 = 0$ are negative. The range of values of $m$ that satisfies this is....

$m < -2 \text{ or } m > -1$

$-2 < m < -1$

$-2 < m < -\frac{1}{2}$

$-2 < m \leq -\frac{7}{16}$

$-1 < m \leq -\frac{7}{16}$

Given the quadratic equation $(m + 2)x^2 - (2m - 1)x + m + 1 = 0$ with the condition that $x_1 < 0$ and $x_2 < 0$ .

For the equation to be a quadratic equation, the coefficient of $x^2$ must not be zero, that is $m + 2 \neq 0$ or $m \neq -2$ .

The discriminant condition is $D \geq 0$ , then we obtain

D = b^2 - 4ac \geq 0

[-(2m - 1)]^2 - 4(m + 2)(m + 1) \geq 0

(2m - 1)^2 - 4(m + 2)(m + 1) \geq 0

4m^2 - 4m + 1 - 4(m^2 + 3m + 2) \geq 0

4m^2 - 4m + 1 - 4m^2 - 12m - 8 \geq 0

-16m - 7 \geq 0

-16m \geq 7

m \leq -\frac{7}{16}

Condition for sum of roots: $x_1 + x_2 < 0$

x_1 + x_2 = -\frac{b}{a} = -\frac{-(2m - 1)}{m + 2} = \frac{2m - 1}{m + 2}

\frac{2m - 1}{m + 2} < 0

The inequality $\frac{2m - 1}{m + 2} < 0$ is satisfied if the numerator and denominator have different signs. Then

(2m - 1)(m + 2) < 0

This inequality is satisfied for $-2 < m < \frac{1}{2}$ .

Condition for product of roots: $x_1x_2 > 0$

x_1x_2 = \frac{c}{a} = \frac{m + 1}{m + 2}

\frac{m + 1}{m + 2} > 0

The inequality $\frac{m + 1}{m + 2} > 0$ is satisfied if the numerator and denominator have the same sign. Then

(m + 1)(m + 2) > 0

This inequality is satisfied for $m < -2$ or $m > -1$ .

Now we combine all conditions:

The intersection of all these conditions is $-1 < m \leq -\frac{7}{16}$ .

Therefore, the values of $m$ that satisfy are $-1 < m \leq -\frac{7}{16}$ .

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