If the equations $\displaystyle ax^{2}+bx+c=0$ and $cx^{2}+bx+a=0,:a\neq c$ have a negative common root, then the value of $a-b+c$ is
- $0$
- $2$
- $1$
-
none of these
Let alpha be the common root. Then a*alpha^2 + b*alpha + c = 0 and c*alpha^2 + b*alpha + a = 0. Subtracting gives (a-c)alpha^2 + (c-a) = 0, so (a-c)(alpha^2 - 1) = 0. Since a != c, alpha^2 = 1, so alpha = 1 or -1. Given the root is negative, alpha = -1. Substituting into the first equation: a(-1)^2 + b(-1) + c = 0, which gives a - b + c = 0.
Let the negative common root be alpha. Applying the condition for a common root, (c times c minus a times a) times alpha squared plus (b times c minus b times a) times alpha plus (a times c minus c times a) equals 0, which simplifies to c squared minus a squared times alpha squared plus b times c minus a times alpha equals 0. Factoring this gives alpha times the quantity c minus a times alpha times c plus a plus b equals 0. Since alpha is not zero and c is not equal to a, the term in the second parentheses must be zero, giving b equals negative alpha times the quantity a plus c. Substituting this expression for b into the expression a minus b plus c gives a plus c plus alpha times a plus c, which factors to a plus c times 1 plus alpha. Since alpha is a negative common root, it must satisfy both equations, and dividing the original equations shows alpha equals negative 1. Therefore, 1 plus alpha is zero, making a minus b plus c equal to 0.