Multiple choice

If $\left( 1-p \right) $ is a root of quadratic equation ${ x }^{ 2 }+px+\left( 1-p \right) =0$, then its roots are

  1. $0, 1$
  2. $-1, 1$
  3. $0, -1$
  4. $-1, 2$
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C Correct answer
Explanation

If (1-p) is a root, it satisfies the equation: (1-p)^2 + p(1-p) + (1-p) = 0. Factoring out (1-p), we get (1-p) * (1-p + p + 1) = 0, which implies (1-p) * 2 = 0, so p = 1. Substituting p = 1 into the original equation gives x^2 + x = 0, so x(x+1) = 0. The roots are 0 and -1.

AI explanation

Substitute x = 1 - p into the quadratic equation x^2 + px + (1 - p) = 0 to get (1 - p)^2 + p(1 - p) + (1 - p) = 0. Factoring out (1 - p) yields (1 - p)(1 - p + p + 1) = 0, which simplifies to 2(1 - p) = 0, giving p = 1. Substituting p = 1 back into the original equation results in x^2 + x = 0, whose roots are found by factoring as x(x + 1) = 0 to be 0 and -1.

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