Multiple choice

Let $p, q$ be real numbers. If $\alpha$ is the root of ${ x }^{ 2 }+3{ p }^{ 2 }x+5{ q }^{ 2 }=0, \beta$ is a root of ${ x }^{ 2 }+9{ p }^{ 2 }x+15{ q }^{ 2 }=0$ and $0 < \alpha < \beta$, then the equation ${ x }^{ 2 }+6{ p }^{ 2 }x+10{ q }^{ 2 }=0$ has a root $\gamma $ that always satisfies

  1. $\gamma =\dfrac { \alpha }{ 4 } +\beta $
  2. $\beta <\gamma $
  3. $\gamma =\dfrac { \alpha }{ 2 } +\beta $
  4. $\alpha <\gamma <\beta $
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D Correct answer
AI explanation

Because alpha is a root of x^2 + 3p^2x + 5q^2 = 0, we have alpha^2 = -3p^2 alpha - 5q^2, and since beta is a root of x^2 + 9p^2x + 15q^2 = 0, we have beta^2 = -9p^2 beta - 15q^2. Define the function f(x) = x^2 + 6p^2x + 10q^2 for the third equation. Evaluating f at alpha yields alpha^2 + 6p^2 alpha + 10q^2, which substitutes to 3p^2 alpha + 5q^2 and is positive since 0 < alpha < beta. Evaluating f at beta yields beta^2 + 6p^2 beta + 10q^2, which substitutes to -3p^2 beta - 5q^2, making it negative because beta is positive. Since f(alpha) > 0 and f(beta) < 0, the intermediate value theorem guarantees the root gamma lies strictly between alpha and beta.

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