Multiple choice

The quadratic polynomials defined on real coefficients $\displaystyle P\left( x \right) ={ a }{ 1 }{ x }^{ 2 }+2{ b }{ 1 }x+{ c }{ 1 }$ $\displaystyle Q\left( x \right) ={ a }{ 12 }{ x }^{ 2 }+2{ b }{ 2 }x+{ c }{ 2 }$ where $\displaystyle { a }{ 1 }\neq 0,\quad { a }{ 2 }\neq 0$ and $\displaystyle P\left( x \right) $ and $\displaystyle Q\left( x \right) $ both take positive values $\displaystyle \forall \times \in Rg\left( x \right) ={ a }{ 1 }{ a }{ 2 }{ x }^{ 2 }+{ b }{ 1 }{ b }{ 2 }x+{ c }{ 1 }{ c }{ 2 }$ then

  1. $g(x)$ takes positive value only
  2. $g(x)$ takes negative value only
  3. $g(x)$ takes both positive and negative value
  4. nothing can be said about $g(x)$
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A Correct answer
Explanation

For each polynomial to be positive for every real x, we have a1 > 0, a2 > 0, c1 > 0, c2 > 0, and b1^2 < a1c1, b2^2 < a2c2. Therefore, (b1b2)^2 < (a1a2)(c1c2), so the quadratic g(x) has a positive minimum and is positive for every real x.

AI explanation

For a quadratic polynomial to be positive for all real x, its leading coefficient must be positive and its discriminant must be negative. Since P(x) = a1*x^2 + 2b1*x + c1 is always positive, we have a1 > 0 and b1^2 - a1*c1 < 0, which means 4b1^2 - 4a1c1 < 0. Applying the same logic to Q(x) gives a2 > 0 and 4b2^2 - 4a2c2 < 0. To evaluate g(x) = a1*a2*x^2 + b1*b2*x + c1*c2, we calculate its discriminant, which is (b1*b2)^2 - (a1*a2)*(c1*c2). Since 4b1^2 < 4a1c1 and 4b2^2 < 4a2c2, multiplying these inequalities yields 16b1^2*b2^2 < 16a1*a2*c1*c2, meaning (b1*b2)^2 - a1*a2*c1*c2 < 0. Because the leading coefficient a1*a2 is the product of two positive numbers, it is also positive, proving that g(x) takes positive values only.

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