Multiple choice

If the equations $ x^2 + ax + bc = 0 $ and $ x^2 + bx + ca = 0$ have a common root and if $a, b $ and $c$ are non-zero distinct real numbers, then their other roots satisfy the equation :

  1. $ x^2 + x + abc = 0 $
  2. $ x^2 - (a+b) x + ab = 0 $
  3. $ x^2 + (a+b) x + ab = 0 $
  4. $ x^2 + x+ ab = 0 $
  5. $ x^2 + abx+ abc = 0 $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If x^2 + ax + bc = 0 and x^2 + bx + ca = 0 have a common root 'k', then k^2 + ak + bc = k^2 + bk + ca, which implies k(a-b) = c(a-b). Since a != b, k = c. Substituting k=c into the first equation: c^2 + ac + bc = 0, so c(c + a + b) = 0. Since c != 0, c = -(a+b). The roots of the first equation are c and b, and the second are c and a. The other roots are a and b, which are roots of x^2 - (a+b)x + ab = 0.

AI explanation

Assume the common root is c. Substituting x = c into x^2 + ax + bc = 0 gives c^2 + ac + bc = 0. Factoring out c, we get c(c + a + b) = 0. Since a, b, and c are distinct non-zero numbers, c + a + b must be 0, making c the common root. To find the remaining root of the first equation, x^2 + ax + bc = 0, we use the sum of roots: r1 + c = -a, so r1 = -a - c. By replacing c with -(a+b), we find r1 = -a - (-(a+b)) = b. Similarly, for the second equation x^2 + bx + ca = 0, the sum of roots is r2 + c = -b, meaning r2 = -b - c. Substituting c = -(a+b) yields r2 = -b - (-(a+b)) = a. A quadratic equation with roots a and b is x^2 - (a+b)x + ab = 0.

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