Multiple choice

Roots of the equation ${x^3} - (a + b + c){x^2} + (ab + bc + ca)x - abc = 0$, if $x^2+2x+7=0$ and $ax^2+bx+c=0$ have a common root, where a,b,c$ \in $R, can be

  1. $4 , 8 , 28$
  2. $1 , 2 , 7$
  3. $1 , 4 , 36$

  4. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The cubic equation factors to (x-a)(x-b)(x-c) = 0, so the roots are a, b, and c. If x^2 + 2x + 7 = 0 and ax^2 + bx + c = 0 share a root, and a, b, c are coefficients, the roots of the quadratic must be the roots of the cubic. By comparing coefficients, the roots are 1, 2, 7.

AI explanation

Let the common root be alpha, meaning alpha^2 + 2alpha + 7 = 0 and a*alpha^2 + b*alpha + c = 0. Since the first equation has complex roots and a, b, and c are real, the second equation must be a scalar multiple of the first, allowing us to write a = k, b = 2k, and c = 7k. The required cubic equation then becomes x^3 - (k + 2k + 7k)x^2 + (2k^2 + 14k^2 + 7k^2)x - 14k^3 = 0. Dividing the entire equation by k^3 and substituting y = x/k yields y^3 - 10y^2 + 23y - 14 = 0, which factors into (y-1)(y-2)(y-7) = 0. Thus, the roots of the given equation are 1, 2, and 7.

Keep practicing — related questions