Question

The number of elements in the set $\left\{\right(a,b)|2a^2+3b^2=35,a,b\in Z\}$ is

• A. $2$
• B. $4$
• C. $8$
• D. $12$

Answer 1

The correct option is C $8$

Let $S=\left\{\right(a,b)|2a^2+3b^2=35,a,b\in Z\}$

$2a^2+3b^2=35$
Putting the value of $a=0,\pm 1,\pm 2,\pm 3,\pm 4$
The values of $b=\pm \sqrt{\frac{35}{3}},\pm \sqrt{11},\pm 3,\pm \sqrt{\frac{17}{3}},\pm 1$

Since $b\in Z$
$\therefore b=\pm 3,\pm 1$
$\therefore S=\left\{\right(2,3),(2,-3),(-2,3),(-2,-3),(4,1),(4,-1),(-4,1),(-4,-1\left)\right\}$
$\therefore$ Total number of ordered pairs $=2\times 2+2\times 2=8$