JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations
Let \(p, q\) and \(r\) be real numbers \((p \ne q,r \ne 0),\) such that the roots of the equation \(\frac{1}{{x + p}} + \frac{1}{{x + q}} = \frac{1}{r}\) are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to .
- A \({p^2} + {q^2} + {r^2}\)
- B \({p^2} + {q^2}\)
- C \(2({p^2} + {q^2})\)
- D \(\frac{{{p^2} + {q^2}}}{2}\)
Answer & Solution
Correct Answer
(B) \({p^2} + {q^2}\)
Step-by-step Solution
Detailed explanation
\(\frac{1}{x+p}+\frac{1}{x+q}=\frac{1}{r}\) \(\frac{{x + p + x + q}}{{(x + p)(x + q)}} = \frac{1}{r}\) \((2x + p + q)r = {x^2} + px + qx + pq\) \({x^2} + (p + q - 2r)x + pq - pr - qr = 0\) Let \(\alpha\) and \(\beta\) be the roots.…
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