What are the solutions of the equation x^4-9x+8=0? Use u substitution to solve

x1 = 8 ^ 1/2

x2 = - 8 ^ 1/2

x3 = 1

x4 = - 1

Step-by-step explanation:

If our equation has the following form: x ^ 4 - 9 * x + 8 = 0, it cannot be solved by substitution, but by factorization:

x ^ 4-9x + 8 = (x - 1) * (x ^ 3 + x ^ 2 + x - 8)

But if it has the form of x ^ 4 - 9x ^ 2 + 8 = 0, if you can:

let u = x ^ 2, then

u ^ 2 - 9 * u + 8 = 0

If we factor, two numbers that added to - 9 but their multiplication is + 8, would be - 8, - 1

- 8 - 1 = - 9

- 8 * - 1 = 8

Thus:

(u - 8) * (u - 1) = 0

(u - 8) = 0 = > u = 8

(u - 1) = 0 = > u = 1

But we know that u is x ^ 2, therefore:

x ^ 2 = 8, x1 = 8 ^ 1/2; x2 = - 8 ^ 1/2

x ^ 2 = 1 x3 = 1, x4 = - 1