To find the real or imaginary solutions of the polynomial equation x^4 - 52x^2 + 576 = 0, we can perform a change of variables:
Let y = x^2, then the equation becomes y^2 - 52y + 576 = 0.
We can now solve this quadratic equation for y. Using the quadratic formula, we have:
y = (-b ± √(b^2 - 4ac)) / (2a)
where a = 1, b = -52, and c = 576.
Plugging in these values, we get:
y = (-(-52) ± √((-52)^2 - 4(1)(576))) / (2(1))
= (52 ± √(2704 - 2304)) / 2
= (52 ± √400) / 2
= (52 ± 20) / 2
This gives us two possible values for y:
1) y = (52 + 20) / 2 = 36
2) y = (52 - 20) / 2 = 16
Now, we substitute these values of y back into y = x^2 to find the possible values for x:
1) For y = 36, we have x^2 = 36. Taking the square root of both sides, we get x = ±√36 = ±6. Therefore, x = 6 or x = -6.
2) For y = 16, we have x^2 = 16. Taking the square root of both sides, we get x = ±√16 = ±4. Therefore, x = 4 or x = -4.
Therefore, the real or imaginary solutions of the polynomial equation x^4 - 52x^2 + 576 = 0 are x = 6, x = -6, x = 4, and x = -4.
what are the real or imaginary solutions of the polynomial equation?
x^4-52x^2+576=0
1 answer