Multiple choice

If $\displaystyle p\,\varepsilon \, Q$ and the quadratic equations $\displaystyle x^{2}-4x+1=0$ and $\displaystyle px^{2}-(p^{2}+3)x+2p^{2}-p=0$ have a root in common, then the value(s) of $p$ are

  1. $1$
  2. $0$
  3. $2$
  4. $-1$
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A Correct answer
Explanation

If x is a common root, x^2 - 4x + 1 = 0. From the second equation, p(x^2 - 4x + 1) + (4px - px - p^2x - 3x + 2p^2 - p) = 0. This simplifies to x(3p - p^2 - 3) + 2p^2 - p = 0. Substituting x^2 = 4x - 1, we find p=1 satisfies the condition.

AI explanation

The condition for two quadratic equations a1x^2 + b1x + c1 = 0 and a2x^2 + b2x + c2 = 0 to have exactly one common root is (c1a2 - c2a1)^2 = (b1c2 - b2c1)(a1b2 - a2b1). Substituting the coefficients gives ((1)(p) - (2p^2 - p)(1))^2 = ((-4)(2p^2 - p) - (-(p^2 + 3))(1))((1)(-(p^2 + 3)) - (p)(-4)). This simplifies to (2p - 2p^2)^2 = (-8p^2 + 4p + p^2 + 3)(-p^2 - 3 + 4p), which further reduces to 4p^2(1 - p)^2 = (p - 1)^2(7p^2 - 10p + 3). Solving this yields valid integer values of p such as 1, making the result 1.

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