The sum of two numbers is 8, and the sum of their squares is 34. What is the larger number?
m+n=8
m^2+n^2=34
n=8-m
n^2=64-16m+m^2
m^2+n^2=34
m^2+64-16m+m^2=34
2m^2-16m+30=0
m^2-8m+15=0
(m-5)(m-3)=0
the two numbers are 3,5
To find the larger number, we need to set up a system of equations based on the given information.
Let's assume the two numbers are x and y, where x is the larger number.
From the first piece of information, we know that x + y = 8. This equation gives us the sum of the two numbers.
From the second piece of information, we know that x^2 + y^2 = 34. This equation gives us the sum of the squares of the two numbers.
To solve this system of equations, we can use substitution or elimination.
Let's solve it using substitution:
From the first equation, we can solve it for x: x = 8 - y.
Substitute this value of x into the second equation:
(8 - y)^2 + y^2 = 34.
Expanding and simplifying:
64 - 16y + y^2 + y^2 = 34.
2y^2 - 16y + 64 - 34 = 0.
2y^2 - 16y + 30 = 0.
Divide by 2 to simplify the equation:
y^2 - 8y + 15 = 0.
Now, we can factorize this quadratic equation:
(y - 5)(y - 3) = 0.
So, y = 5 or y = 3.
Now, substitute these values of y back into the equation x + y = 8 to find the corresponding values of x.
For y = 5:
x + 5 = 8,
x = 8 - 5,
x = 3.
For y = 3:
x + 3 = 8,
x = 8 - 3,
x = 5.
Therefore, the larger number is 5.