TS EAMCET · Maths · Quadratic Equation
If both roots of the equation \(x^2-5 a x+6 a=0\) exceed 1, then the range of ' \(a\) ' is
- A \([-1,0) \cup\left[\frac{24}{25}, \infty\right)\)
- B \(\left[\frac{24}{25}, \infty\right)\)
- C \([-1,0)\)
- D \(\mathbb{R}\)
Answer & Solution
Correct Answer
(B) \(\left[\frac{24}{25}, \infty\right)\)
Step-by-step Solution
Detailed explanation
\(\Delta \ge 0 \implies (-5a)^2 - 4(1)(6a) \ge 0 \implies 25a^2 - 24a \ge 0 \implies a(25a - 24) \ge 0 \implies a \in (-\infty, 0] \cup [\frac{24}{25}, \infty)\) \(f(1) > 0 \implies 1^2 - 5a(1) + 6a > 0 \implies 1 + a > 0 \implies a > -1\) Axis of symmetry…
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