For what values of m, the equation mx2 - (m + l)x + 2m - 1 = 0 does not possesses any real roots.

Clearly, D < 0(m+1)2−4m(2m−1)<07m2−6m−1<0(m−1)(7m+1)<0m∈−17,1

For what values of m, the equation mx2 - (m + l)x + 2m - 1 = 0 does not possesses any real roots.

Clearly, D < 0(m+1)2−4m(2m−1)<07m2−6m−1<0(m−1)(7m+1)<0m∈−17,1

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For what values of m, the equation mx² - (m + l)x + 2m - 1 = 0 does not possesses any real roots.

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$- \begin{aligned} m \in (- \frac{1}{7}, 1) \end{aligned}$

$\begin{aligned} m \in (- \frac{1}{8}, 1) \end{aligned}$

$\begin{aligned} m \in (\frac{1}{7}, 1) \end{aligned}$

$\begin{aligned} m \in (- \frac{1}{7}, 1) \end{aligned}$

answer is D.

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Clearly, D < 0

$\begin{array}{l} (m + 1)^{2} - 4 m (2 m - 1) < 0 \\ 7 m^{2} - 6 m - 1 < 0 \\ (m - 1) (7 m + 1) < 0 \\ m \in (- \frac{1}{7}, 1) \end{array}$

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For what values of m, the equation mx2 - (m + l)x + 2m - 1 = 0 does not possesses any real roots.

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For what values of m, the equation mx² - (m + l)x + 2m - 1 = 0 does not possesses any real roots.