if r is a root, then (x-r) is a factor
(x-4)(x-7) = x^2 - 11x + 28
x^2 - 11x + 28 > 0
the parabola crosses the x-axis at 4 and 7
... function values outside of the roots are greater than zero
The solution below corresponds to an inequality involving a quartic function.
Write a possible quartic polynomial inequality that matches the solution. Prove that your inequality matches the solution.
x < 4 and x > 7
4 answers
but i need a quartic equation
you know that x^4 > 0 has solutions
x < 0 or x > 0
So, since 11/2 is midway between 4 and 7, just shift right and you have
4-11/2 = 3/2
7-11/2 = 3/2
so,
(x - 11/2)^4 > (3/2)^4
see
http://www.wolframalpha.com/input/?i=(x+-+11%2F2)%5E4+%3E+(3%2F2)%5E4
and scroll down to where it shows the solution set
x < 0 or x > 0
So, since 11/2 is midway between 4 and 7, just shift right and you have
4-11/2 = 3/2
7-11/2 = 3/2
so,
(x - 11/2)^4 > (3/2)^4
see
http://www.wolframalpha.com/input/?i=(x+-+11%2F2)%5E4+%3E+(3%2F2)%5E4
and scroll down to where it shows the solution set
Thanks a lot