Multiple choice

If both the roots of the equation $x^{2}+mx+1=0$ and $(b-c)x^{2}+(c-a)x+(a-b)=0$ are common then:

  1. $m=-2$
  2. $m=-1$
  3. $m=0$
  4. $m=1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If two quadratic equations have the same roots, their coefficients must be proportional: 1/(b-c) = m/(c-a) = 1/(a-b). This implies b-c = a-b, so 2b = a+c. Also, m = (c-a)/(b-c). Since a-b = b-c, c-a = -2(b-c). Thus m = -2(b-c)/(b-c) = -2.

AI explanation

By the factor theorem, the equation (b - c)x squared + (c - a)x + (a - b) = 0 must have x = 1 as a root because the sum of its coefficients is zero. Therefore, x = 1 is also a root of x squared + m(x) + 1 = 0. Substituting 1 into this equation gives 1 + m + 1 = 0, so m is -2.

Keep practicing — related questions