Question

how would find the x-intercept of f(x)=(1/4)x^4-2x^2+1=0 how would find the x-intercept of f(x)=(1/4)x^4-2x^2+1=0 @Mathematics

Answer 1

The x intercept of any equation is where y=0. So you'd want to make y (or in this case f(x)) 0 and solve.

Answer 2

solve the quadratic equation
\[\frac{1}{4}z^2-2z+1=0\] and once solved, replace z by
\[x^2\]

Answer 3

And that would be how you'd solve ^

Answer 4

if it were me i would start with
\[z^2-8z+4=0\] then
\[z^2-8z=-4\]
\[(z-4)^2=-4+16=12\]
\[z-4=\pm\sqrt{12}\]
\[z=4\pm2\sqrt{3}\]

Answer 5

also what is the y and x intercepts for x*sqrt(2-x^(2))

Answer 6

not done yet, because both these values are positive. you then have to solve
\[x^2=4+2\sqrt{3}\] and
\[x^2=4-2\sqrt{3}\]

Answer 7

but if you want a snappier way you can write
\[ (x^2-2 x-2) (x^2+2 x-2)=0\] and then solve two quadratic equation separately

Answer 8

how did you get rid of the 1/4

Answer 9

i can remove it if i have an equation. this is set = 0 so i can multiply by 4 and it is still = 0

Answer 10

how did you get 8z+4?

Answer 11

i didn't mean
\[f(x) = (x^2-2 x-2) (x^2+2 x-2)\] because
\[f(x) = \frac{1}{4} (x^2-2 x-2) (x^2+2 x-2)\]

Answer 12

also what is the y and x intercepts for x*sqrt(2-x^(2))