Question 1109304: For how many integers x is the number x^4 - 51x^2 + 50 negative?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the equation is negative between -sqrt(50) and -1, and between 1 and sqrt(50).
sqrt(50) is equal to 7.071 rounded to 3 decimal places.
therefore, the equation is negative between -7.071 and -1, and between 1 and 7.071.
it is 0 at -7.071 and 0 at -1 and 0 at 1 and 0 at 7.071, therefore it is not negative at those values of x.
it is negative from -7 to -2, and from 2 to 7, inclusive.
that would be 12 integers total.
you find the 0 points by factoring the equation.
this is a fourth degree equation that can be solved as if it is a quadratic equation.
the equation is x^4 -51x^2 + 50 = 0 in standard form.
let z = x^2 and it becomes z^2 - 51z + 50 = 0
factor this quadratic equation to get (z-50) * (z-1) = 0
solve for z to get z = 50 or z = 1
since z = x^2, you get x^2 = 50 or x^2 = 1
solve for x to get x = plus or minus sqrt(50) or x = plus or minus 1.
those are the zero crossing points of x^4 - 51x^2
+ 50
now you check the intervals between all these zero crossing points.
you will find that the graph starts off positive until it gets to -sqrt(50), then goes negative until it gets to -1, then goes positive until it gets to 1, then goes negative until it gets to sqrt(50), then goes positive the rest of the way.
the graph of the equation looks like this:
from the graph, you can see that the equation is negative between -7.071 and -1, and it is negative between 1 and 7.071.
7.071 is the square root of 50 rounded to 3 decimal places.
there are 6 integers between -7.071 and -1, not including -7.071 and -1.
they are -7, -6, -5, -4, -3, -2.
there are 6 integers between 1 and 7.071, not including 1 and 7.071.
they are 2, 3, 4, 5, 6, 7.
that's 12 integers total.
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