If $a\in$Z and the equation $(x-a)(x-10)+1=0$ has integral roots, then the values of $a$ are
- $10, 8$
- $12, 10$
- $12, 8$
- $18, 10$
(x-a)(x-10) = -1. For integral roots, (x-a) and (x-10) must be factors of -1. Possible pairs: (1, -1) or (-1, 1). Case 1: x-a = 1, x-10 = -1 => x = 9, a = 8. Case 2: x-a = -1, x-10 = 1 => x = 11, a = 12.
Expanding the equation (x-a)(x-10)+1=0 gives x^2 - (a+10)x + 10a + 1 = 0. For this equation to have integral roots, the product of the roots must be 10a + 1. We can rewrite the expanded equation as x^2 - ax - 10x + 10a = -1. Factoring by grouping gives x(x-10) - a(x-10) = -1, which simplifies to (x-10)(x-a) = -1. For x and a to be integers, the only factor pairs of -1 are 1 and -1 or -1 and 1. If x-10 = 1 and x-a = -1, then x = 11 and a = 12. If x-10 = -1 and x-a = 1, then x = 9 and a = 8. The values of a are 12 and 8.