Question

If \(a\in \mathbb{Z}\) and the equation \[ (x-a)(x-10)+1=0 \] has integral roots, then the values of \(a\) are:

• A. \(8,12\)
• B. \(12,10\)
• C. \(10,8\)
• D. \(9\)

Hint:

Whenever a quadratic equation is required to have integral roots, first compute the discriminant and force it to be a perfect square.

Answer 1

The Correct Option is A

Concept: If a quadratic equation has integral roots, then its discriminant must be a perfect square. For a quadratic equation \[ x^2-Sx+P=0, \] the roots are integers only when \[ D=S^2-4P \] is a perfect square.

Step 1: Expand the given equation.
\[ (x-a)(x-10)+1=0 \] \[ x^2-(a+10)x+10a+1=0 \]

Step 2: Calculate the discriminant.
\[ D=(a+10)^2-4(10a+1) \] \[ =a^2+20a+100-40a-4 \] \[ =a^2-20a+96 \] \[ =(a-10)^2-4 \]

Step 3: Set the discriminant equal to a perfect square.
Let \[ (a-10)^2-4=m^2 \] Then \[ (a-10)^2-m^2=4 \] \[ (a-10-m)(a-10+m)=4 \] Possible factor pairs of \(4\): \[ (2,2),\ (1,4),\ (-1,-4),\ (-2,-2) \] Solving gives \[ a=12,\quad a=8 \] \[ \boxed{a=8,\;12} \]