Question

Number of different values of $p$ for which the equation $x^2-2(p+q)x+(q^2+2qp)=0$ has integral roots is $(p,q\in N,p\in [5,10])$

• A. $6$
• B. $4$
• C. $0$
• D. $5$

Answer 1

The correct option is A

$6$

The roots of the equation are $\frac{2(p+q)\pm \sqrt{\left(2\right(p+q){)}^2-4(q^2+2pq)}}{2}$

$=\frac{2(p+q)\pm \sqrt{4(p+q{)}^2-4(q^2+2pq)}}{2}=\frac{2(p+q)\pm 2\sqrt{p^2+q^2+2pq-q^2-2pq}}{2}=\frac{2(p+q)\pm 2p}{2}=p+q\pm p$

$\Rightarrow$ The roots will be an integer for all values of $p\in [5,10]$ and $p\in N$.
Hence, number of value of $p$ which satisfies our condition is $6$.