Multiple choice

If A, G & H are respectively the AM, GM and HM of three positive numbers a, b & c then the equation whose roots are a, b & c is given by

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

For roots a, b, c, the equation is x^3 - (a+b+c)x^2 + (ab+bc+ca)x - abc = 0. Given AM = (a+b+c)/3 = A, GM^3 = abc = G^3, and HM = 3abc/(ab+bc+ca) = H, we have ab+bc+ca = 3G^3/H. Substituting gives x^3 - 3Ax^2 + 3(G^3/H)x - G^3 = 0.

AI explanation

For any cubic polynomial with roots a, b, and c, the polynomial can be written using Vieta's formulas as x^3 - (a+b+c)x^2 + (ab+bc+ca)x - abc = 0. The sum of the roots equals 3A, and the product of the roots equals G^3. The sum of the pairwise products ab+bc+ca can be rewritten using the relationship between the arithmetic, geometric, and harmonic means of three numbers as abc * (1/a + 1/b + 1/c) = G^3 * (3/H). Substituting these equivalents into the standard polynomial form yields x^3 - 3Ax^2 + 3(G^3/H)x - G^3 = 0.

Keep practicing — related questions