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Let $p,q,r,s$ be real numbers such that $pr=2(q+s)$. Consider the equations $x^2+px+q=0$ and $x^2+rx+s=0$. Then

  1. at least one of the equations has real roots
  2. both these equations have real roots
  3. neither of these equations have real roots
  4. given data is not sufficient to arrive at any conclusion
in Numerical Ability by Boss (17.5k points)
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1 Answer

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The discriminant of $x^2+px+q=0$ is $Δ_1=p^2−4q$

and that of $x^2+rx+s=0$ is $Δ_2=r^2−4s$

and $Δ_1+Δ_2=p^2−4q+r^2−4s$

$= p^2+r^2−4(q+s)$

$= (p+r)^2−2pr−4(q+s)$

$= (p+r)^2−2[pr−2(q+s)]$

and if $pr=2(q+s)$, we have $Δ_1+Δ_2=(p+r)^2$

Since, the sum of two discriminants is positive, so at least one of them has to be positive.

Therefore, at least one the equation $x^2+px+q  = 0 $ and $x^2+rx + s = 0 $ has the real roots.

by Boss (19.1k points)
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