Question

The sum of the squares of the two consecutive positive integers exceed their product by $91.$ Find the integers?

• A. $9,10$
• B. $10,11$
• C. $11,12$
• D. $12,13$

Answer 1

The correct option is A $9,10$

Let the two consecutive positive integers be $x$ and $x+1$

According to the given condition

$x^2+(x+1{)}^2=91+x(x+1)$

$\Rightarrow x^2+(x+1{)}^2-x(x+1)=91$

$\Rightarrow x^2+x^2+1+2x-x^2-x=91$

$\Rightarrow x^2+x-90=0$

$\Rightarrow (x+10)(x-9)=0$

$\Rightarrow x=-10$ or $x=9$.

As $x$ is a positive integer,

$\therefore x=9$

Hence, the two consecutive positive integers are $9$ and $10$.