Answer:
1. 3, -3, 4i, -4i.
2. 1, √5i, - √5i.
Explanation:
1. 7x^2 - 144 = -x^4
x^4 + 7x^2 - 144 = 0
Let y = x^2, then we have:-
y^2 + 7y - 144 = 0
(y + 16)(x - 9) = 0
y = -16 or 9.
So x +/- √9 or +/- √-16
x = 3, -3, 4i, -4i (answer).
2. x^3-x^2+5x-5=0
f(1) = 1 - 1 + 5 - 5 = 0
Therefore x = 1 is a zero.
Therefore x - 1 is a factor of the function.
To find the other zeroes we divide the function by x - 1:-
x^2 + 5 <----- is the result.
--------------------------
x - 1 ) x^3 - x^2 + 5x - 5
x^3 - x^2
5x - 5
5x - 5.
So we have:-
x^2 + 5 = 0
x^2 = -5
x = +/- √5 i.