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JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations

यदि \(p\) तथा \(q\) दो धनात्मक संख्याएँ है, जिनके लिए \(p+q=2\) तथा \(p^{4}+q^{4}=272\) है, तो \(p\) तथा \(q\) जिस समीकरण के मूल है, वह है

  1. A \(x^{2}-2 x+2=0\)
  2. B \(x^{2}-2 x+8=0\)
  3. C \(x^{2}-2 x+136=0\)
  4. D \(x^{2}-2 x+16=0\)
Verified Solution

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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