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JEE Main 2022 · Mathematics · Quadratic Equations · medium

If α,β are the roots of x^2-3x-5=0, then the quadratic equation whose roots are α+(1)/(β) and β+(1)/(α) is

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If α,β\alpha,\beta are the roots of x2−3x−5=0x^2-3x-5=0, then the quadratic equation whose roots are α+1β\alpha+\dfrac1\beta and β+1α\beta+\dfrac1\alpha is
  1. A. 5x2+12x−16=05x^2+12x-16=0
  2. B. 5x2−12x−16=05x^2-12x-16=0
  3. C. 5x2−12x+16=05x^2-12x+16=0
  4. D. x2−12x−16=0x^2-12x-16=0
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Answer: (B) 5x2−12x−16=05x^2-12x-16=0

Here α+β=3\alpha+\beta=3 and αβ=−5\alpha\beta=-5. The sum of the new roots is (α+β)+(1α+1β)=(α+β)+α+βαβ=3−35=125(\alpha+\beta)+\left(\frac1\alpha+\frac1\beta\right)=(\alpha+\beta)+\frac{\alpha+\beta}{\alpha\beta}=3-\frac35=\frac{12}{5}. Their product is (α+1β)(β+1α)=αβ+1+1+1αβ=−5+2−15=−165\left(\alpha+\frac1\beta\right)\left(\beta+\frac1\alpha\right)=\alpha\beta+1+1+\frac1{\alpha\beta}=-5+2-\frac15=-\frac{16}{5}. The required equation is x2−125x−165=0x^2-\frac{12}{5}x-\frac{16}{5}=0, that is 5x2−12x−16=05x^2-12x-16=0. Its roots 44 and −45-\frac45 can be cross-checked against the transformed roots.
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