Question
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.
The sum of the second number and the square of the first is 115.
Answers
Scott
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
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
Bosnian
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