The sum of the squares of two numbers is 2. The product of the two numbers is 1. Find the numbers.
xy=1
x^2+y^2=2
x^4+1=2x^2
x^4-2x^2+1=0
I don't think you can factor that
unless its (x^2+1)(x^2-1)
but I don't know where to go from there
***********************
x^4-2x^2+1=0
I don't think you can factor that
unless its (x^2+1)(x^2-1)
***********************
If you FOIL that, you will get x^4 - 1, not x^4 - 2x^2 + 1.
You can factor by grouping
x^4 - 2x^2 + 1 = 0
x^4 - (x^2 + x^2) + 1 = 0
x^4 - x^2 - x^2 + 1 = 0
x^2(x^2 - 1) + -1(x^2 - 1) = 0
(x^2 - 1)(x^2 - 1) = 0
x^2 = 1
x = 1
To solve the equation x^4 - 2x^2 + 1 = 0, you can make a substitution to simplify the equation. Let's use the substitution u = x^2.
Substituting u = x^2 into the equation, we get:
u^2 - 2u + 1 = 0
Now, we can try factoring this quadratic equation:
(u - 1)(u - 1) = 0
(u - 1)^2 = 0
Setting each factor equal to zero, we have:
u - 1 = 0
u = 1
Now, substitute back x^2 for u:
x^2 = 1
Taking the square root of both sides:
x = ±1
Therefore, the two numbers that satisfy the given conditions are x = 1 and x = -1.
To find the values of x and y, we can solve the system of equations given:
1. xy = 1
2. x^2 + y^2 = 2
Substituting the value of y from equation 1 into equation 2, we get:
x^2 + (1/x)^2 = 2
x^2 + 1/x^2 = 2
Now, let's multiply through by x^2 to get rid of the fraction:
(x^2)(x^2) + (1/x^2)(x^2) = 2x^2
x^4 + 1 = 2x^2
Rearranging the equation:
x^4 - 2x^2 + 1 = 0
At this point, we can try to factor this equation. Notice that this equation is a quadratic equation in terms of x^2. Let's substitute x^2 with a new variable, let's say u:
u^2 - 2u + 1 = 0
This quadratic equation can be factored as:
(u - 1)^2 = 0
So, we have:
u - 1 = 0
u = 1
Now, let's substitute u back in terms of x:
x^2 = u
x^2 = 1
x = ±√1
Therefore, the possible values for x are +1 and -1.
Using equation 1, we can find the corresponding values of y:
For x = 1, y = 1/1 = 1
For x = -1, y = 1/-1 = -1
Hence, the two numbers that satisfy the given conditions are (1, 1) and (-1, -1).