If the equation (m - 3)x^2 - 2mx + 5m = 0 has real and positive roots, then the possible value(s) of m lies in: Option 1: [2,3] Option 2: (2,3.8) Option 3: (3, 3.75] Option 4: [3.1, 4)

Answer: (3, 3.75]. Solution: Hints : The roots are real and positive if D≥0, sum of roots ≥0 and product of roots > 0. (I) D≥0 4m^2-20m (m-3) ≥ 0 -4m^2 + 15m ≥ 0 4m^2-15m ≤ 0 m [0,3.75] (II) arrow Sum of roots >0: 2m/(m-3) >0 m>0 m (-∞,0) (3,∞) (III) arrow product of roots >0: 5m/(m-3) >0 m/(m-3) >0 m (-∞,0) (3,∞) Combining (I), (II) and (III) Hence m (3,3.75]

Algebra

Quadratic Equations

Hard

If the equation $(m - 3)x^2 - 2mx + 5m = 0$ has real and positive roots, then the possible value(s) of m lies in:

[2,3]

(2,3.8)

(3, 3.75]

[3.1, 4)

✅ Correct Option: 3

Hints :

The roots are real and positive if $D\ge0$ , sum of roots $\ge0$

and product of roots $> 0$ .

(I) $D\ge0 \implies 4m^2-20m (m-3) \ge 0 \implies -4m^2 + 15m \ge 0$

$\implies 4m^2-15m \le 0 \implies m \in [0,3.75]$

(II) $\rightarrow$ Sum of roots $>0$ :

$\frac{2m}{m-3} >0 \implies m>0 \quad m \in (-\infty,0) \cup (3,\infty)$

(III) $\rightarrow$ product of roots $>0$ :

$\frac{5m}{m-3} >0 \implies \frac{m}{m-3} >0 \implies m \in (-\infty,0) \cup (3,\infty)$

Combining (I), (II) and (III)

Hence $m \in (3,3.75]$

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BBA-DBE Bangalore 2026

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If the equation (m−3)x2−2mx+5m=0(m - 3)x^2 - 2mx + 5m = 0(m−3)x2−2mx+5m=0 has real and positive roots, then the possible value(s) of m lies in:

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BBA-DBE Bangalore Past Year Questions (Topic-Wise)

If the equation $(m - 3)x^2 - 2mx + 5m = 0$ has real and positive roots, then the possible value(s) of m lies in: