Multiple choice

If the equation $ { x^{ 2 } }-2px+q=0$ has two equal roots, then the equation $ \left( 1+y \right) { x }^{ 2 }-2\left( p+y \right) x+\left( q+y \right) =0 $ will have its roots real and distinct only when

  1. $y$ is not negative and $p$ is not unity.
  2. $y$ is negative and $p$ is not unity.
  3. $y$ is negative and $p$ is unity.

  4. None of these.
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B Correct answer
AI explanation

If x^2 - 2px + q = 0 has equal roots, its discriminant gives q = p^2. Substituting q = p^2 into the second equation and analyzing its discriminant yields 4(p - 1)^2 - 4y. For the second equation to have real and distinct roots, 4(p - 1)^2 - 4y > 0. This means (p - 1)^2 > y. If y is negative and p is not unity, this inequality is guaranteed to hold. The result is that y is negative and p is not unity.

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