JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations
Let \(a, b, c\) be the length of three sides of a triangle satisfying the condition \(\left(a^2+b^2\right) x^2-2 b(a+c)\). \(x+\left(b^2+c^2\right)=0\). If the set of all possible values of \(x\) is the interval \((\alpha, \beta)\), then \(12\left(\alpha^2+\beta^2\right)\) is equal to ...........
- A \(30\)
- B \(36\)
- C \(35\)
- D \(37\)
Answer & Solution
Correct Answer
(B) \(36\)
Step-by-step Solution
Detailed explanation
\(\left(a^2+b^2\right) x^2-2 b(a+c) x+b^2+c^2=0 \) \( \Rightarrow a^2 x^2-2 a b x+b^2+b^2 x^2-2 b c x+c^2=0\) \( \Rightarrow(a x-b)^2+(b x-c)^2=0 \) \( \Rightarrow a x-b=0, \quad b x-c=0 \) \( \Rightarrow a+b>c \quad b+c>a \quad \quad c+a>b\) \(a+a x>b x \) \(a+a x > a x^2 \)…
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