JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations
The minimum value of the sum of the squares of the roots of \(x^{2}+(3-a) x+1=2 a\) is.
- A \(4\)
- B \(5\)
- C \(6\)
- D \(8\)
Answer & Solution
Correct Answer
(C) \(6\)
Step-by-step Solution
Detailed explanation
\(\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta\) let \(f(a)=(3-a)^{2}-2(1-2 a)\) \(f ( a )= a ^{2}-2 a +7\) \(f ( a )=( a -1)^{2}+6\) \(f ( a ))_{\min .}=6\)
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