Answer:
First integer = 3
Second Integer = 4
Step-by-step explanation:
Let:
First integer = x
Second Integer = x + 1
Now according to given conditions:
3 x^2 = 7 + 5 (x + 1)
3 x^2 = 7 + 5x + 5
3 x^2 = 12 + 5x
Transferring all terms to left side and changing their signs
3 x^2 - 5x - 12 = 0
By quadratic formula:
a = 3
b = -5
c = -12
x = [tex]\frac{(-b)±\sqrt{b^2-4ac} }{2a}[/tex]
By putting values of a, b and c we get
x = -1 and 3
By ignoring negative integer, take x = 3
Now:
First integer = 3
Second Integer = 4
Proof:
3 x^2 = 7 + 5 (x + 1)
Putting x = 3
3* 3^2 = 7 + 5 (3 + 1)
3* 9 = 7 + 5 (4)
27 = 7 + 20
27 = 27
Hence proved
I hope it will help you!
