If the equation $x^{2}+bx-a=0$ and $x^{2}-ax+b=0$ have a root in common then the value of $a-b$ will be

Let alpha be the common root of x^2 + bx - a = 0 and x^2 - ax + b = 0. By substituting alpha into both equations, we get alpha^2 + b*alpha - a = 0 and alpha^2 - a*alpha + b = 0. Subtracting the second equation from the first eliminates alpha^2, yielding b*alpha + a*alpha - a - b = 0. Factoring out alpha from the first two terms gives alpha(a + b) = a + b. For this to hold true, either a + b = 0 or alpha = 1. If a + b = 0, then a = -b, which implies a - b = -2b, but substituting a = -b into the first equation gives x^2 + bx + b = 0, requiring b to be specific values that do not universally yield an integer result. If alpha = 1, substituting it back into the first equation gives 1 + b - a = 0. Rearranging this equation gives a - b = 1. Therefore, the value of a - b is 1.

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