Multiple choice

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

  1. $a=6$ and $14$
  2. $a=8$ and $12$
  3. $a=10$ and $10$

  4. none
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B Correct answer
Explanation

The equation is x^2 - (a+10)x + 10a + 1 = 0. For roots to be integers, the discriminant D = (a+10)^2 - 4(10a+1) must be a perfect square. D = a^2 + 20a + 100 - 40a - 4 = a^2 - 20a + 96 = (a-10)^2 - 4. Let (a-10)^2 - 4 = k^2. Then (a-10)^2 - k^2 = 4. The only squares differing by 4 are 4 and 0. So (a-10)^2 = 4, meaning a-10 = 2 or -2. Thus a = 12 or 8.

AI explanation

Expand the given equation to x squared minus (a + 10)x plus (10a + 1) equals 0. For the roots to be integers, the discriminant, (a + 10) squared minus 4(10a + 1), must be a perfect square, which simplifies to a squared minus 20a plus 96. Factoring this expression gives (a minus 8)(a minus 12), which equals zero when a is 8 or 12. Therefore, the integer values of a are 8 and 12.

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