Let's first solve for x1 and x2 using the given system of equations:
x1 + x2 = 5 (1)x1*x2 = -3 (2)
From equation (1), we can rewrite it as x1 = 5 - x2 and substitute it into equation (2):
(5 - x2) * x2 = -3
Expanding:5x2 - x2^2 = -3Rearranging:x2^2 - 5x2 - 3 = 0
Using the quadratic formula:x2 = [5 ± √(5^2 - 4*(-3))]/2x2 = [5 ± √(25 + 12)] / 2x2 = [5 ± √37] / 2
So the solutions for x2 are:x2 = (5 + √37) / 2 or x2 = (5 - √37) / 2
Now, let's find x1 using x2 = (5 + √37) / 2:x1 = 5 - x2x1 = 5 - (5 + √37) / 2x1 = 10/2 - (5 + √37) / 2x1 = (10 - 5 - √37) / 2x1 = (5 - √37) / 2
Therefore, the solutions for x1 and x2 are:x1 = (5 - √37) / 2x2 = (5 + √37) / 2
Now, we can find x1^4 + x2^4:x1^4 + x2^4 = [(5 - √37) / 2]^4 + [(5 + √37) / 2]^4
Calculating these values would yield the final result for x1^4 + x2^4.
Let's first solve for x1 and x2 using the given system of equations:
x1 + x2 = 5 (1)
x1*x2 = -3 (2)
From equation (1), we can rewrite it as x1 = 5 - x2 and substitute it into equation (2):
(5 - x2) * x2 = -3
Expanding:
5x2 - x2^2 = -3
Rearranging:
x2^2 - 5x2 - 3 = 0
Using the quadratic formula:
x2 = [5 ± √(5^2 - 4*(-3))]/2
x2 = [5 ± √(25 + 12)] / 2
x2 = [5 ± √37] / 2
So the solutions for x2 are:
x2 = (5 + √37) / 2 or x2 = (5 - √37) / 2
Now, let's find x1 using x2 = (5 + √37) / 2:
x1 = 5 - x2
x1 = 5 - (5 + √37) / 2
x1 = 10/2 - (5 + √37) / 2
x1 = (10 - 5 - √37) / 2
x1 = (5 - √37) / 2
Therefore, the solutions for x1 and x2 are:
x1 = (5 - √37) / 2
x2 = (5 + √37) / 2
Now, we can find x1^4 + x2^4:
x1^4 + x2^4 = [(5 - √37) / 2]^4 + [(5 + √37) / 2]^4
Calculating these values would yield the final result for x1^4 + x2^4.