If the product of two numbers is 65 and one of the numbers is 8 more than the other, what is the remaining number?
What are the factors of 65?
TRY DIVIDING 65 BY 15
Lily -- that doesn't work.
65 / 15 = 4.333333333
sorry i meant 13
To find the remaining number, we can set up a system of equations based on the given information.
Let's assume that one of the numbers is "x" and the other number is "x + 8" since one number is 8 more than the other.
According to the problem, the product of these two numbers is 65. This can be represented as:
x * (x + 8) = 65
Now, we can solve this equation to find the value of "x" and then find the value of the remaining number.
Expanding the equation:
x^2 + 8x = 65
Rearranging the equation to make it equal to zero:
x^2 + 8x - 65 = 0
Now, we need to solve this quadratic equation to find the values of "x". We can use the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / (2a)
In this equation, a = 1, b = 8, and c = -65.
Plugging in the values:
x = (-(8) ± sqrt((8)^2 - 4(1)(-65))) / (2(1))
Simplifying further:
x = (-8 ± sqrt(64 + 260)) / 2
x = (-8 ± sqrt(324)) / 2
x = (-8 ± 18) / 2
Now, we have two possible values for "x". Let's calculate each one separately:
For x = (-8 + 18) / 2:
x = 10 / 2
x = 5
For x = (-8 - 18) / 2:
x = -26 / 2
x = -13
Both values are possible solutions, but one of them doesn't match the given condition that one number is 8 more than the other. Since the other number is the one that is 8 more, we can discard the negative value.
Therefore, the remaining number is 5.