Question

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

• A. $a=6$ and $14$
• B. $a=8$ and $12$
• C. $a=10$ and $10$
• D. none

Answer 1

The correct option is B $a=8$ and $12$

Given that,

$\begin{array}{c}\left(x-a\right)\left(x-10\right)+1=0\\ x^2-10x-ax+10a+1=0\\ x^2-x\left(a+10\right)+\left(10a+1\right)=0\\ \because b^2-4ac=0\\ {\left(a+10\right)}^2-4\times 1\times \left(10a+1\right)=0\\ a^{2}100+20a-40a-4=0\\ a^2-20a+96=0\\ a^2-12a-8a+96=0\\ a\left(a-12\right)-8\left(a-12\right)=0\\ \left(a-12\right)\left(a-8\right)=0\end{array}$

$a=12$ or $a=8$

Hence,

Option $\left(B\right)$ is correct answer