The two real solutions of the equation x⁴ - 4x³ + 4x² + 4x - 5 = 0 are 1 and -1.
How to find the real solutions of the equation?
We have the cubic equation:
x⁴ - 4x³ + 4x² + 4x - 5 = 0
And we want to find the real solutions for this equation.
We can rewrite this as:
x⁴ + 4*(-x³ + x² + x) - 5 = 0
Notice that if x = 1, we have a solution, so x = 1 is a factor.
1⁴ + 4*(-1³ + 1² + 1) - 5 = 0
0 = 0
And also, x = -1 is also a solution
(-1)⁴ + 4*(-(-1)³ +(-1)² + (-1)) - 5 = 0
then we can write the polynomial as:
(x + 1)*(x - 1)*[ax² + bx + c] = x⁴ - 4x³ + 4x² + 4x - 5
Because a is on the quartic term, we know that a = 1
(x + 1)*(x - 1)*[x² + bx + c] = x⁴ - 4x³ + 4x² + 4x - 5
Then c*-1*1 = -5,
c = -5/-1 = 5
so:
(x + 1)*(x - 1)*[x² + bx + -5] = x⁴ - 4x³ + 4x² + 4x - 5
And for b we can write:
bx*x + b*x*x + x²*1 + x²*-1 = 4x²
2bx² = 4x²
2b = 4
b = 4/2
Then the equation becomes:
(x + 1)*(x - 1)*[x² + 2x + 5] = x⁴ - 4x³ + 4x² + 4x - 5
Now we need to find the roots of:
x² + 2x + 5
Using the quadratic formula we get:

We have a negative argument in the square root, so the solutions are not real, and we can discard them.
The only real solutions are 1 and -1.