Question

find all real zeros of the polynomial function
f(x)= -54^4+80x^2

Answer 1

A. x=0, x=+-4
B. x=0, x=16
C. x=0,x=4
D. x=,x=+-16

Answer 2

are you sure that's the right equation? because none of those are zeros of \(80x^2-54^4\)

Answer 3

oh sorry the -54 is a -5

Answer 4

@kirbykirby

Answer 5

none of those are roots still even if that's -5^4

Answer 6

f(x)= -5x^4+80x^2

Answer 7

oh there's an x.. ok

Answer 8

\(-5x^4+80x^2=-5(x^2)^2+80(x^2)=0\) if you substitute \(x^2=y\) you get a regular quadratic \(-5y^2+80y=0\)

Answer 9

\(5y(-y+16)=0\)
This implies \(5y=0~or~-y+16=0\)

Answer 10

Can you solve for y?

Answer 11

idk how

Answer 12

1/16?

Answer 13

5y=0 -> y=0

-y+16=0 .... -y=-16....y=16

Answer 14

but we said \(y=x^2\), so we have
\(x^2=0\) and \(x^2=16\)
then solve for x

Answer 15

32

Answer 16

@kirbykirby

Answer 17

\(x^2=16\) means \(x=-4\) or \(x=4\)

Answer 18

since \((-4)^2=16\) and \(4^2=16\)

Answer 19

So B right/?

Answer 20

A. x=0, x=+-4 B. x=0, x=16 C. x=0,x=4 D. x=,x=+-16

Answer 21

it would be A