I am struggling with the following question and was hoping for some assistance:
How many real solutions does this equation have?: (x^4) - (5x^2) +6 = 0
I inserted the brackets to make it easier to read. The solution in the book reads '4', with no other details whatsoever.
The first step I did was to determine positive and negative case and look for sign changes:
Pos: (x^4) - (5x^2) + 6 = 0
One from x^4 to 5x^2
And one from 5x^2 to 6
Neg: (-x^4) - (-5x^2) + 6 = 0
Which becomes (-x^4) + 5x^2 + 6 = 0
One change: -x^4 to +5x^2.
So the total should be 3 correct? I have tried doing this several times and can't figure out what I am missing.
In addition, the Purplemath page on the Rule of Signs has some quirks that are confusing me, copied below:
Code: Select all
Now I look at f(–x) (that is, having changed the sign on x, so this is the "negative" case):
f (–x) = (–x)^5 – (–x)^4 + 3(–x)^3 + 9(–x)^2 – (–x) + 5
= –x^5 – x^4 – 3x^3 + 9x^2 + x + 5
Shouldn't the negative case read = -x^5 plus x^4 - 3x^3 minus 9x^2 + x + 5?
Actually, I just had a eureka moment: Is it the -X being squared/cubed that is affecting the signs?
= -x^5 - (-x)^4
= -x^5 - (-x * -x * -x * -x)
= -x^5 - (x^2 * x^2)
= -x^5 - (x^4)
And thus there is not actually a double-negative sign and the term remains a subtraction?