If the three equations $x^{2}+ax+12=0,x^{2}+bx+15=0,x^{2}+(a+b)x+36=0$ have a common positive root then $a$ and $b$ are

Let the common positive root be k, so it satisfies all three equations, giving k^2 + ak + 12 = 0, k^2 + bk + 15 = 0, and k^2 + (a+b)k + 36 = 0. Adding the first two equations gives 2k^2 + (a+b)k + 27 = 0, and subtracting the third equation from this result yields k^2 - 9 = 0. Because the root is positive, k = 3, which we substitute back into the first two equations to get 9 + 3a + 12 = 0 and 9 + 3b + 15 = 0. Solving these gives a = -7 and b = -8.

• Thanks
• Wrong explanation