s = 2 f - 5
s + f^2 = 115
f^2 + 2 f - 5 = 115
f^2 + 2 f - 120 = 0
(f + 12)(f - 10) = 0
f - 10 = 0 ... f = 10
Determine two positive integers such as 5 less than two times the first number is the second number.
The sum of the second number and the square of the first is 115.
2 answers
a = first number
b = second number
b = 2 a - 5
b + a ^ 2 = 115
2 a - 5 + a ^ 2 = 115
- 5 + a ^ 2 + 2 a = 115 Subtract 115 to both sides
- 5 + a ^ 2 + 2 a - 115 = 115 - 115
- 5 + a ^ 2 + 2 a - 115 = 0
Try to slove this equation.
The solutions are :
a = - 12 and a = 10
- 12 isn't positive integer so a = 10
b = b = 2 a - 5 = 2 * 10 - 5 = 20 -5 = 15
b = second number
b = 2 a - 5
b + a ^ 2 = 115
2 a - 5 + a ^ 2 = 115
- 5 + a ^ 2 + 2 a = 115 Subtract 115 to both sides
- 5 + a ^ 2 + 2 a - 115 = 115 - 115
- 5 + a ^ 2 + 2 a - 115 = 0
Try to slove this equation.
The solutions are :
a = - 12 and a = 10
- 12 isn't positive integer so a = 10
b = b = 2 a - 5 = 2 * 10 - 5 = 20 -5 = 15
1 answer