Multiple choice

Consider two quadratic expressions $f(x)=ax^2+bx+c$ and $g(x)=ax^2+px+q$, $(b\neq q)$ such that their discriminants are equal. If $f(x)=g(x)$ has a root $x=\alpha$, then?

  1. $\alpha$ will be A.M. of the roots of $f(x)=0$ and $g(x)=0$
  2. $\alpha$ will be A.M. of the roots of $f(x)=0$
  3. $\alpha$ will be A.M. of the roots of $g(x)=0$

  4. None of these
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A Correct answer
Explanation

Given f(x) = ax^2 + bx + c and g(x) = ax^2 + px + q, f(x) = g(x) implies (b-p)x + (c-q) = 0, so x = (q-c)/(b-p). With equal discriminants, b^2 - 4ac = p^2 - 4aq, which simplifies to b^2 - p^2 = 4a(c-q). Factoring gives (b-p)(b+p) = 4a(c-q), so (c-q)/(b-p) = -(b+p)/4a. The roots of f(x)=0 are (-b +/- sqrt(D))/2a and g(x)=0 are (-p +/- sqrt(D))/2a. The A.M. of all four roots is (-b-p)/4a, which matches alpha.

AI explanation

Equating f(x) and g(x) gives ax squared plus bx plus c equals ax squared plus px plus q, which simplifies to (b minus p)x equals (q minus c), yielding the root alpha equals (q minus c) divided by (b minus p). From the equality of the discriminants, b squared minus 4ac equals p squared minus 4aq, which rearranges to 4a(q minus c) equals (b minus p)(b plus p). Dividing this relationship by 4a times (b minus p) shows that alpha equals (b plus p) divided by 2a, which is exactly half the sum of the axes of symmetry of the two parabolas and thus the arithmetic mean of the roots of both f(x) equals 0 and g(x) equals 0.

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