JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations
Let \(p\) and \(q\) be two positive numbers such that \(p + q =2\) and \(p ^{4}+ q ^{4}=272 .\) Then \(p\) and \(q\) are roots of the equation
- A \(x^{2}-2 x+2=0\)
- B \(x^{2}-2 x+8=0\)
- C \(x^{2}-2 x+136=0\)
- D \(x^{2}-2 x+16=0\)
Answer & Solution
Correct Answer
(D) \(x^{2}-2 x+16=0\)
Step-by-step Solution
Detailed explanation
Consider \(\left(p^{2}+q^{2}\right)^{2}-2 p^{2} q^{2}=272\) \(\left((p+q)^{2}-2 p q\right)^{2}-2 p^{2} q^{2}=272\) \(16-16 p q+2 p^{2} q^{2}=272\) \((p q)^{2}-8 p q-128=0\) \(( pq )^{2}-8 pq -128=0\) \(pq =\frac{8 \pm 24}{2}=16,-8\) \(\therefore \quad pq =16\)…
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