Two consecutive positive integers have a product of 7832?

Question

Two consecutive positive integers have a product of 7832. Find the numbers.

Answer 1

Let the numbers be x and x + 1

x(x+1) = 7832

x^2 + x - 7832 = 0

By the quadratic formula

x = (-1 +/- sqrt(1 - 4(1)(-7832))) / 2
= (-1 +/- sqrt(1 + 31328)) / 2
= (-1 +/- 177) / 2
= 176 / 2 or -178/2
= 88 or -89

We want POSITIVE integers, so x = 88 is our solution.

The numbers are 88 and 89

CHECK
88 * 89 = 7832

Answer 2

n(n+1)=7832

n^2 + n - 7832 = 0 use quad formula

b^2 - 4ac = 1 + 4(7832) = 31329

sqrt(31329) = 177

so n =( -1 +/- 177 )/2 = -89 or 88. 88 is the positive one.

88 * 89 = 7832

Answer 3

If the integers are n and n+1, n2+n = 7832 and
you can factor that to get n. Probably 88*89

Answer 4

x (x + 1) =7832
x² + x - 7832 = 0
(x - 88) (x + 89) = 0
Accept x = 88 (positive)
Integers are 88 and 89

Answer 5

Let the Two Consecutive positive integers be n and n+1 Given n*(n+1)=306 n^2+n-306=0 Factorize, n^2+18n-17n-306=0 n(n+18)-17(n+18)=0 n-17=0; n+18=0 Since n is +ve integer we can only take n=17

Answer 6

a*b=7832

a-b=1 -------- a=1+b

(b+1)*b = 7832
b^2 +b = 7832
b^2 +b - 7832 = 0
factorising
(x-88)(x+89)=0

so the values that you are looking for are 88 and 89
88*89=7832

Answer 7

x = 1st integer, x + 1 = 2nd integer

x(x + 1) = 7,832
x^2 + x - 7,832 = 0
(x + 89)(x - 88) = 0
1st integer: x - 88 = 0, x = 88
2nd integer: = 88 + 1 or 89

Answer: 88 and 89 are the integers.

Proof:
= 88 * 89
= 7,832