Multiple choice

The equations ${ x }^{ 3 }+5{ x }^{ 2 }+px+q=0$ and ${ x }^{ 3 }+7{ x }^{ 2 }+px+r=0$ have two roots in common. If the third root of each equation is represented by ${x}{1}$ and ${x}{2}$ respectively, then the ordered pair $({x}{1},{x}{2})$ is:

  1. $(-5,-7)$
  2. $(1,-1)$
  3. $(-1,1)$
  4. $(5,7)$
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A Correct answer
Explanation

Subtracting the two equations: (x^3+5x^2+px+q) - (x^3+7x^2+px+r) = 0 => -2x^2 + q - r = 0. This doesn't help directly. Let roots be a, b, x1 and a, b, x2. Sum of roots: a+b+x1 = -5 and a+b+x2 = -7. Thus x1 - x2 = 2. Also product of roots: abx1 = -q and abx2 = -r. The common roots satisfy x^2(x+5+x1) = 0? No, use Vieta's. The common roots a, b satisfy a+b = -5-x1 and a+b = -7-x2. Also ab + a(x1) + b(x1) = p and ab + a(x2) + b(x2) = p. So ab + (a+b)x1 = ab + (a+b)x2 => (a+b)(x1-x2) = 0. Since x1 != x2, a+b = 0. Then x1 = -5 and x2 = -7.

AI explanation

Let the two common roots be m and n, so the first equation can be written as (x - x1)(x - m)(x - n) = 0 and the second as (x - x2)(x - m)(x - n) = 0. Expanding these, the first gives x^3 - (x1+m+n)x^2 + ... = 0, so x1 + m + n = -5. The second equation gives x2 + m + n = -7. Subtracting the first equation's expanded form from the second gives (x3^3 - x1^3) + 2(x3^2 - x1^2) + (r - q) = 0, but an easier approach is subtracting the original equations to get 2x^2 + (r - q) = 0. Substituting this back reveals the non-common roots must balance the difference in the x^2 coefficients, yielding x1 = -5 and x2 = -7.

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