Find the smallest of three consecutive positive integers if the product of the smaller two integers is 2 more than 8 times the largest integer.

Question

asked Sep 11, 2023 58.9k views

1 Answer

Brnunes · Answer 1 · 2023-09-18T04:17:08+0000

Answer:

9

Explanation:

Let the three consecutive positive integers be:

If the product of the two smaller integers is 2 more than 8 times the largest integer then:

$\implies x(x+1)=8(x+2)+2$

Solve the equation for x:

$\implies x^2+x=8x+16+2$

$\implies x^2+x=8x+18$

$\implies x^2-7x-18=0$

$\implies x^2-9x+2x-18=0$

$\implies x(x-9)+2(x-9)=0$

$\implies (x+2)(x-9)=0$

Therefore:

$\implies x+2=0 \implies x=-2$

$\implies x-9=0 \implies x=9$

As x is a positive integer, x = 9.