AP EAMCET · Maths · Quadratic Equation
Let \(a, b\) and \(c\) be three positive real numbers such that the sum of any two of them is greater than the third. All the values of \(\lambda\) such that the roots of the equation \(x^2+2(a+b+c) x+3 \lambda(a b+b c+c a)=0\) are real, are given by
- A \(\lambda < \frac{2}{3}\)
- B \(\lambda \geq \frac{2}{3}\)
- C \(\lambda < \frac{4}{3}\)
- D \(\frac{1}{3} < \lambda < \frac{2}{3}\)
Answer & Solution
Correct Answer
(C) \(\lambda < \frac{4}{3}\)
Step-by-step Solution
Detailed explanation
Roots of the equation \[ x^2+2(a+b+c) x+3 \lambda(a b+b c+c a)=0 \] are real, so \[ \begin{aligned} & {[2(a+b+c)]^2-4(1)[3 \lambda(a b+b c+c a)] \geq 0} \\ & \Rightarrow \quad \lambda \leq \frac{a^2+b^2+c^2+2(a b+b c+c a)}{3(a b+b c+c a)} \end{aligned} \]…
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