Multiple choice

If $A, G, H$ be respectively A.M., GM., H.M. of three numbers (>0) then the equation whose roots are the numbers is given by

  1. $\displaystyle x^{3}+3Ax^{2}+3\frac{G}{H}x-G^{3}=0$
  2. $\displaystyle x^{3}-3Ax^{2}-3(\frac{G^3}{H})+G^{3}=0$
  3. $x^{3}-3Ax^{2}+G^{3}\left ( 3x-1 \right )=0$
  4. $\displaystyle x^{3}-3Ax^{2}+3(\frac{G^3}{H})x-G^{3}=0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For roots a, b, c, the cubic is x^3 - (a+b+c)x^2 + (ab+bc+ca)x - abc = 0. Given A = (a+b+c)/3, G^3 = abc, and H = 3/(1/a+1/b+1/c) = 3abc/(ab+bc+ca). Thus, ab+bc+ca = 3abc/H = 3G^3/H. Substituting gives x^3 - 3Ax^2 + 3(G^3/H)x - G^3 = 0.

AI explanation

Let the three numbers be the roots of a cubic equation, where their sum is 3A. The product of the roots is G^3, which serves as the constant term. The sum of the products of the roots taken two at a time equals 3(G^3)/H, giving the coefficient of x. Therefore, the cubic equation is x^3 - 3Ax^2 + 3(G^3/H)x - G^3 = 0.

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