[tex]Solve \ x^4 - 17x^2 + 16 = 0[/tex]
To solve this equation , we need to write it in quadratic form
[tex]ax^2+bx+c=0[/tex]
To get the equation in quadratic form we replace x^2 with u
[tex]x^2 = u[/tex]
[tex] x^4 - 17x^2 + 16 = 0[/tex] can be written as
[tex](x^2)^2 - 17x^2 + 16 = 0[/tex], Replace u for x^2
So equation becomes
[tex]u^2 - 17u + 16 = 0[/tex]
Now we factor the left hand side
-16 and -1 are the two factors whose product is +16 and sum is -17
(u-16) (u-1) = 0
u -16 = 0 so u=16
u-1 =0 so u=1
WE assume u = x^2, Now we replace u with x^2
[tex]u = 16 so \ it \ becomes \ x^2 =16[/tex]
Now take square root on both sides , x= +4 and x=-4
[tex]u = 1 so \ it \ becomes \ x^2 =1[/tex]
Now take square root on both sides , x= +1 and x=-1
So zeros of the function are -4, -1, 1, 4
The zeros of the equation are 4 and -4
Given the polynomial expression x^4 – 17x^2 + 16 = 0.
Reduce the function to a quadratic function
Let P = x²
The equation becomes;
(x²)² – 17x² + 16 = 0
P² - 17P + 16 =0
Factorize the result
P² - 16P - P+ 16 =0
P(P-16)-1(P-16) = 0
(P-16) (P-16) = 0
P = 16 twice
Since P = x²
x² = 16
x = ±√16
x = ±4
Hence the zeros of the equation are 4 and -4
Learn more here: https://brainly.com/question/12182022
