Question

The product of two consecutive odd integers is 63. What are the integers? Use a quadratic equation to solve for the integers.

Answer 1

Let n, n+2, and n+4 represent the three consecutive even integers.

Answer 2

The integers are odd. Let the integers are

$2n+1, 2n+3$

And the given product is 63 .

That is

$(2n+1)(2n+3) = 63 \\ 4n^2 + 6n +2n + 3 -63=0 \\ 4n^2 +8n -60=0$

Dividing whole equation by 4

$n^2 + 2n -15=0 \\ (n+5)(n-3) =0$

$n+5=0, n-3=0 \\ n=-5, n=3$

When n = -5, numbers are

$2(-5)+1 , 2(-5)+3 = -9, -7$

When n =3, we will get

$2(3)+1 , 2(3) + 3 = 7, 9$