Two whole numbers add to 17 and the sum of their squares is 185. What are the numbers
follow the same logic I used in your previous post:
x+y = 18
x^2+y^2 = 185
go for it
except, don't make my typo. Better say
x+y = 17
not real numbers
17^2 = 289 which is greater than 185
anyway
a^2 + (17-a)^2 = 185
a^2 + 289 - 34 a + a^2 = 185
2 a^2 - 34 a + 104 = 0
a^2 -17 a + 52 = 0
(a- 13)(a-4) = 0
a = 4 or 13
so they are 4 and 13 and my initial thought was wrong
To find the two whole numbers, we can set up a system of equations. Let's call the two numbers x and y.
From the given information, we have two equations:
Equation 1: x + y = 17 (The sum of the two numbers is 17)
Equation 2: x^2 + y^2 = 185 (The sum of their squares is 185)
We can solve this system of equations to find the values of x and y.
Let's solve using the substitution method:
From Equation 1, we can rewrite it as y = 17 - x.
Now we substitute this value of y in Equation 2:
x^2 + (17 - x)^2 = 185
Expanding and simplifying, we get:
x^2 + 289 - 34x + x^2 = 185
Combining like terms, we have:
2x^2 - 34x + 104 = 0
To solve this quadratic equation, we can either factor or use the quadratic formula. In this case, factoring is not straightforward, so let's use the quadratic formula:
x = (-b +/- sqrt(b^2 - 4ac)) / (2a)
Using the quadratic formula, with a = 2, b = -34, and c = 104, we get:
x = (-(-34) +/- sqrt((-34)^2 - 4 * 2 * 104)) / (2 * 2)
Simplifying further, we have:
x = (34 +/- sqrt(1156 - 832)) / 4
x = (34 +/- sqrt(324)) / 4
x = (34 +/- 18) / 4
Now we have two possible values of x:
1. x = (34 + 18) / 4 = 52 / 4 = 13
2. x = (34 - 18) / 4 = 16 / 4 = 4
Now that we have the possible values of x, we can substitute them back into Equation 1 to find the corresponding values of y.
If x = 13, then y = 17 - 13 = 4
If x = 4, then y = 17 - 4 = 13
Therefore, the two whole numbers that add to 17 and have a sum of squares equal to 185 are 4 and 13.