The square of the sum of two numbers is 144, and the sum of their squares is 80. Find the number
Let's assume the two numbers are x and y. We are given two pieces of information:
1) The square of the sum of the two numbers is 144: (x + y)^2 = 144
Expanding this equation, we get x^2 + 2xy + y^2 = 144.
2) The sum of their squares is 80: x^2 + y^2 = 80.
We can solve this system of equations by subtracting equation 2 from equation 1 to eliminate the y^2 term:
x^2 + 2xy + y^2 - (x^2 + y^2) = 144 - 80
2xy = 64
xy = 32
Now let's use this value of xy in either equation 1 or equation 2 to find the values of x and y.
Let's use equation 2:
x^2 + y^2 = 80
x^2 + (32/x)^2 = 80
Rearranging this equation, we get x^4 - 80x^2 + 1024 = 0.
We can factor this equation as (x^2 - 64)(x^2 - 16) = 0.
This gives us two equations:
1) x^2 - 64 = 0
x^2 = 64
x = ±8
2) x^2 - 16 = 0
x^2 = 16
x = ±4
So the numbers are x = 8, y = 4 or x = -8, y = -4.
Let's assume the two numbers as "x" and "y".
Based on the given problem, we have two equations:
1. (x + y)^2 = 144
2. x^2 + y^2 = 80
To solve these equations step-by-step, let's expand the equation (x + y)^2:
1a. (x + y) * (x + y) = 144
Expanding (x + y) * (x + y):
1b. x^2 + xy + yx + y^2 = 144
Simplifying the equation:
1c. x^2 + 2xy + y^2 = 144
Now we have the following two equations:
1c. x^2 + 2xy + y^2 = 144
2. x^2 + y^2 = 80
To find the values of x and y, we can subtract equation 2 from equation 1c:
1c - 2:
(x^2 + 2xy + y^2) - (x^2 + y^2) = 144 - 80
Simplifying the equation:
2xy = 64
Now, let's divide the equation by 2:
xy = 32
We have found the value of xy as 32.
Now, we need to find the values of x and y that satisfy the given conditions.
To find the values of x and y, we can solve this equation as a system of equations using substitution or elimination.
Let's solve it using substitution:
From equation 2, we have:
x^2 + y^2 = 80
Rearranging the equation:
y^2 = 80 - x^2
Substituting the value of xy from equation 1:
32 = yx
Rearranging the equation:
y = 32 / x
Substituting the value of y in the equation y^2 = 80 - x^2:
(32 / x)^2 = 80 - x^2
Expanding the equation:
(32^2 / x^2) = 80 - x^2
Multiplying both sides by x^2:
32^2 = 80x^2 - x^4
Simplifying the equation:
1024 = 80x^2 - x^4
Rearranging the equation:
x^4 - 80x^2 + 1024 = 0
Now we have a quadratic equation in terms of x.
We can solve this equation by factoring or using the quadratic formula.
Factoring the equation is a bit complicated in this case, so let's use the quadratic formula:
The quadratic formula is given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 1, b = -80, and c = 1024.
Using the quadratic formula:
x = (-(-80) ± √((-80)^2 - 4(1)(1024))) / (2(1))
Simplifying the equation:
x = (80 ± √(6400 - 4096)) / 2
x = (80 ± √(2304)) / 2
x = (80 ± 48) / 2
Now, we have two possible values for x:
x = (80 + 48) / 2 = 128 / 2 = 64
or
x = (80 - 48) / 2 = 32 / 2 = 16
Thus, the possible values for x are 64 and 16.
Now, let's find the corresponding values of y:
Substituting x = 64 in the equation xy = 32:
64 * y = 32
y = 32 / 64
y = 0.5
So, one possible set of numbers is x = 64 and y = 0.5.
Substituting x = 16 in the equation xy = 32:
16 * y = 32
y = 32 / 16
y = 2
Another possible set of numbers is x = 16 and y = 2.
In conclusion, the two possible sets of numbers that satisfy the given conditions are:
1. x = 64 and y = 0.5
2. x = 16 and y = 2.
To find the numbers, let's denote the first number as "x" and the second number as "y."
We are given two conditions:
1) The square of the sum of the two numbers is 144:
(x + y)^2 = 144
2) The sum of their squares is 80:
x^2 + y^2 = 80
To solve this system of equations, we can use the method of substitution.
Let's begin by simplifying the first equation:
(x + y)^2 = 144
Expanding the equation, we have:
x^2 + 2xy + y^2 = 144
Now, using the second equation, we can substitute (x^2 + y^2) with 80:
(x^2 + 2xy + y^2) = 80
Replacing (x^2 + y^2) with 80 in the expanded equation, we have:
80 + 2xy = 144
Rearranging this equation and isolating 2xy, we get:
2xy = 144 - 80
2xy = 64
Dividing both sides of the equation by 2, we find:
xy = 32
Now, we have two equations:
1) xy = 32
2) x^2 + y^2 = 80
From equation 1, we can solve for x in terms of y:
x = 32 / y
Substituting this value of x into equation 2, we have:
(32 / y)^2 + y^2 = 80
Expanding and simplifying this equation, we get:
1024 / y^2 + y^2 = 80
Multiplying through by y^2 to eliminate the denominator, we get:
1024 + y^4 = 80y^2
Rearranging this equation, we have:
y^4 - 80y^2 + 1024 = 0
Let's substitute a new variable, z = y^2. The equation becomes:
z^2 - 80z + 1024 = 0
Now we have a quadratic equation in terms of z. We can solve this equation using factoring or the quadratic formula.
Factoring the equation, we have:
(z - 64)(z - 16) = 0
This gives us two possible values for z:
1) z - 64 = 0, which gives z = 64
2) z - 16 = 0, which gives z = 16
Taking the square root of both values of z, we find:
1) z = 64, which gives y^2 = 64 and hence y = ±8
2) z = 16, which gives y^2 = 16 and hence y = ±4
Now, we can substitute these values of y back into equation 1 to find the corresponding values of x:
1) If y = 8, then xy = 32 implies x = 32 / 8 = 4
2) If y = -8, then xy = 32 implies x = 32 / -8 = -4
3) If y = 4, then xy = 32 implies x = 32 / 4 = 8
4) If y = -4, then xy = 32 implies x = 32 / -4 = -8
So, the two numbers are:
1) (4, 8)
2) (-4, -8)
3) (8, 4)
4) (-8, -4)
Therefore, there are four pairs of numbers that satisfy the given conditions: (4, 8), (-4, -8), (8, 4), and (-8, -4).