Question

The number of ordered triplets $(p,q,r)$ of real numbers that satisfy the simultaneous equations
$p+q=2$
$pq-r^2=1$ is

• A. $0$
• B. $1$
• C. $2$
• D. infinitely many

Answer 1

The correct option is B $1$

$pq=1+r^2\ge 1$
i.e., $pq\ge \mathrm{1.......}\left(1\right)$
$(p-q{)}^2=(p+q{)}^2-4pq$
$\Rightarrow (p-q{)}^2=4-4pq\ge 0\Rightarrow pq\le 1...(2)$


From $\left(1\right)$ and $\left(2\right),$ we get
$pq=1\Rightarrow (p-q{)}^2\le 0$
$\Rightarrow p=q$
Thus, the only solution is $p=q=1,r=0$