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JEE Main 2018 · Mathematics · Quadratic Equations · medium

The equations x^2+px+q=0 and x^2+qx+p=0 with p≠ q have exactly one common root. If the two non-common roots differ by 4 in absolute value, then 4pq…

Single correct option

The equations x2+px+q=0x^2+px+q=0 and x2+qx+p=0x^2+qx+p=0 with p≠qp\neq q have exactly one common root. If the two non-common roots differ by 44 in absolute value, then 4pq4pq equals
  1. A. −15-15
  2. B. 1515
  3. C. −9-9
  4. D. −25-25
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Answer: (A) −15-15

Subtracting the equations gives (p−q)x+(q−p)=0(p-q)x+(q-p)=0, so the common root is x=1x=1 (as p≠qp\ne q). Substituting, 1+p+q=01+p+q=0, i.e. p+q=−1p+q=-1. In the first equation the product of roots is qq, so its other root is qq; in the second the product is pp, so its other root is pp. Hence ∣p−q∣=4|p-q|=4. Taking p−q=4p-q=4 with p+q=−1p+q=-1 gives p=32p=\tfrac32, q=−52q=-\tfrac52 (the other sign just swaps them). Then 4pq=4⋅32⋅(−52)=−154pq=4\cdot\tfrac32\cdot\left(-\tfrac52\right)=-15. Check: x2+1.5x−2.5x^2+1.5x-2.5 has roots 1,−2.51,-2.5; x2−2.5x+1.5x^2-2.5x+1.5 has roots 1,1.51,1.5.
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Original question written in the JEE Main 2018 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake