For which values of $m$ does the quadratic equation $3 x^{2} + 2 m x + (m - 5) = 0$ have real and distinct roots?

medium MCQ 2 marks Class 10 · Mathematics · Quadratic Equations

Answer

All real values of

m

The correct answer is All real values of $m$ — Therefore the required set is all real numbers.

Step-by-step solution

Real and distinct roots require the discriminant to be positive: $Δ = b^{2} - 4 a c > 0$ .
Identify $a = 3$ , $b = 2 m$ , $c = m - 5$ .
Compute $Δ = (2 m)^{2} - 4 \times 3 \times (m - 5) = 4 m^{2} - 12 (m - 5) = 4 m^{2} - 12 m + 60$ .
Factor out 4: $Δ = 4 (m^{2} - 3 m + 15)$ .
Examine the quadratic $m^{2} - 3 m + 15$ : its discriminant is $(- 3)^{2} - 4 \times 1 \times 15 = 9 - 60 = - 51 < 0$ , so $m^{2} - 3 m + 15$ is always positive for every real $m$ .
Hence $Δ = 4 \times$ (positive) is always positive, meaning the original quadratic has real and distinct roots for every real $m$ .
Therefore the required set is all real numbers.

Common mistake: Assuming the discriminant can become negative for some

m

without checking the sign of the resulting expression.

Concept tested: Discriminant >0 for real distinct roots