Question

The product of two consecutive positive integers is 11 more than their sum. find integers

Answer 1

We will call N the first consecutive positive integer.

Then

N(N + 1) = N + (N +1) +11

N² +N = N + N +1 +11 = N² +N = 2N +12

N² + N - 2N -12 = 0

N² - N - 12 = 0

$N = \frac{1+-\sqrt{1^{2} + 4*1*12}}{2*1} = \frac{1*-\sqrt{49}}{2} = \frac{1+-7}{2}$

We only need positive integers, then

then N = (1 + 7)/2 = 8/2 = 4

and N + 1 = 4 + 1 = 5

ANSWER two positive consecutive integers are 4 and 5

checking

4*5 = 4+5 +11

20 = 9 +11 = 20 match!!

$\textbf{Spymore}$​

Answer 2

$n(n+1)=n+n+1+11\\n^2+n=2n+12\\n^2-n-12=0\\n^2+3n-4n-12=0\\n(n+3)-4(n+3)=0\\(n-4)(n+3)=0\\n=4 \vee n=-3\\\\-3\not>0\\n=4\\n+1=5$

So, it's 4 and 5.