If the quadratic equations $x^2+abx+c=0$ and $x^2+acx+b=0$ have a common root, the equation containing their other roots is/are:
- $x^2+a(b+c)x-a^2 bc=0$
- $x^2-a (b+c)x+a^2 bc=0$
- $a(b+c) x^2-(b+c)x+abc=0$
- $a(b+c) x^2+(b+c)x-abc=0$
Let alpha be the common root. alpha^2+ab*alpha+c=0 and alpha^2+ac*alpha+b=0. Subtracting: a(b-c)*alpha + (c-b) = 0. a(b-c)*alpha = b-c. If b!=c, alpha = 1/a. Substituting alpha=1/a into the first equation: 1/a^2 + b + c = 0, so 1+a^2(b+c)=0. The product of roots for eq1 is c, so the other root is c/alpha = ac. For eq2, product is b, other root is b/alpha = ab. The equation with roots ac and ab is x^2 - (ac+ab)x + a^2bc = 0, or x^2 - a(b+c)x + a^2bc = 0.
Substituting x = -a into the first equation gives a^2 - a^2 b + c = 0, meaning c = a^2(b - 1). The second equation gives a^2 - a^2 c + b = 0, meaning b = a^2(c - 1). Multiplying these results in bc = a^4(b - 1)(c - 1), which simplifies to a^2 = bc because a^2(b - 1) = c and a^2(c - 1) = b. The other roots are b/a and c/a, and their required quadratic equation is x^2 - (b/a + c/a)x + (bc/a^2) = 0. Substituting a^2 = bc gives the equation x^2 - a(b + c)x + a^2 bc = 0.