Find two consecutive integers such that their product is 19 more than their sum?

Question

I would really appreciate it.

Answer 1

let the integrs be x and x+1
According to the problem :
x(x+1)=x+x+1+19
=>x^2+x=2x+20
=>x^2+x-2x-20=0
=>x^2-x-20=0
=>x^2-5x+4x-20=0
=>x(x-5)+4(x-5)=0
=>(x-5)9x+4)=0
kTherefore either x-5=0 or x+4=0
x=5 or -4
The numbers are 5 and 6 or -4 and -3.

Answer 2

The integers should be denoted as

n and n+1

so the equations are as follows:

n(n+1) = 19 + (n + (n+1) )

n^2 +n = 19 + 2n + 1

n^2 + n = 20 +2n

n^2 -n -20 = 0

(n - 5)(n + 4) = 0

n - 5 = 0 so n = 5

if n = 5 then n+1 = 6

the integers are 5 and 6 are one possible solution

n + 4 = 0 so n = -4

if n = -4 then n+1 = -3

the integers -4 and -3 are also a solution

Answer 3

Given consecutive integers n and n+1:
n(n+1) - 19 = 2n + 1
n^2 + n -19 = 2n + 1
n^2 -n -20 = 0
(n+4)(n-5) = 0
n = -4, 5
(n, n+1) = (-4, -3) or (5, 6)

Answer 4

x=4 and x+1=5,
so it's 4 and 5

Answer 5

I agree with MB

because if you were to plug the numbers back in to check, they equal each other...

[-3, -4]

(-3) * (-4) = 19 + (-3) + (-4)

12 = 16 + (-4)

12 = 12

[6,5]

(6) * (5) = 19 + 6 + 5

30 = 25 + 5

30 = 30

Answer 6

5 and 6. Sum=11, Product=30.

Hope that helps!

Answer 7

x + (x+1) + 19 = x(x+1)
2x + 20 = x^2 + x
wait... no... i am not doing this right... i'm sorry!! i totally forgot how to do this!!