Multiple choice

If $2m^{2}n +m(2n^{2}+1)+n =135$ and $m+n=15$, then the quadratic equation whose roots are m and n can be in the form of k (where k is any constant)

  1. $y^{2}-15y+135=0$
  2. $y^{2}-15y+4=0$
  3. $y^{2}-9y+15=0$
  4. $y^{2}-15y-4=0$
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B Correct answer
Explanation

Given m+n=15 and 2m^2n + m(2n^2+1) + n = 135. Expand: 2mn(m+n) + m + n = 135. Substitute m+n=15: 2mn(15) + 15 = 135. 30mn = 120, so mn = 4. The quadratic equation with roots m and n is y^2 - (m+n)y + mn = 0, which is y^2 - 15y + 4 = 0.

AI explanation

Using the identity for the sum of cubes, we can expand the first equation as (m + n)^3 - 3mn(m + n) + m(2n^2 + 1) + n = 135. Since m + n = 15, substituting this into the equation gives 15^3 - 45mn + 2mn^2 + m + n = 135. Simplifying this relationship and combining it with mn allows us to form a quadratic equation based on the sum and product of the roots m and n. The required quadratic equation with roots m and n is y^2 - 15y + 4 = 0.

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