Multiple choice

Find all the integral values of $a$ for which the quadratic equation $\left ( x-a \right )\left ( x-10 \right )+1=0$ has integral roots.

  1. $12, 8$
  2. $11, 9$
  3. $12, 9$
  4. $11, 8$
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A Correct answer
Explanation

(x-a)(x-10) = -1. For roots to be integers, (x-a) and (x-10) must be factors of -1. Factors are (1, -1) or (-1, 1). Case 1: x-10 = 1 => x=11. Then x-a = -1 => 11-a = -1 => a=12. Case 2: x-10 = -1 => x=9. Then x-a = 1 => 9-a = 1 => a=8.

AI explanation

Expanding the equation gives x squared minus 10 plus a times x plus 10a minus 1 equals 0. Using the condition for integral roots based on perfect squares, 10 minus a plus 1 and a minus 10 plus 1 must be perfect squares, leading to 11 minus a equals 1 and a minus 9 equals 1, so a can be 10 or 12. Alternatively, if 10 minus a minus 1 and a minus 10 minus 1 are perfect squares, then a minus 11 equals 1 and 9 minus a equals 1, giving a as 12 or 8; the only consistent values matching the viable integer pairs for the roots are 12 and 8.

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