I need to know how to simplify:
x^(3/4) + 5/x^(1/4) (you may assume x>0)
And how do I find the absolute inequality corresponding to: x<=-22/3 OR x>=11/3 (And what exactly is the absolute inequality?)
Multiply the first term by x^(1/4)/x^(1/4) :
x^(3/4) * x^(1/4)/x^(1/4) + 5/x^(1/4)
= x^((3/4)+(1/4))/x^(1/4) + 5/x^(1/4)
= x^1/x^(1/4) + 5/x^(1/4)
= x / x^(1/4) + 5/x^(1/4)
= (x+5) / x^(1/4)
Absolute Inequality I think refers to something like |x|<3, therefore -3<x<3, so we need an absolute value (in other words it is the distance from 0, ignoring plus or minus)
x<=-22/3 OR x>=11/3 could possibly be solved by taking the difference of the two absolute values:
|-22/3| - |11/3| = 11/3
Then halve that and put it in a formula: |x+11/6|<11/3+11/6 ==> |x+11/6|<11/2
Why did I do this? Because I wanted to know how to find the "middle" of those two numbers. (Try putting dots on the number line to see what I mean).
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
At first I was confused, but when I tried your number line suggestion I got your meaning. Thank you so much for you help! But now I'm having trouble with this problem:
Solve for X:
x^3+2x^2-x=-2 Is this right? If it is then, I can't seem to go any further
x³ + 2x² - 9x = 18
x³ + 2x² - 9x - 18 = 0
Putting x=3, we find that it is one of the solutions.
(x-3)(x² + 5x + 6) = 0
x² + 5x + 6 = (x+3)(x+2)
(x-3)(x+3)(x+2) = 0
which gives the following values of x :- 3, -3, -2
PS:_ A cubic equation, that is an equation of degree three, can be solved only by trial and error, as far as I know.
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Oh, I came out with a different answer...so is this wrong?:
x^3 + 2x^2 - 9x = 18
x^3 + 2x^2 - 9x-18 = 0
3^3 + 2(3)^2 - 9(3) = 18
What you've done is correct, just incomplete.
For the x²-9=0 bit, that means that x²=9, so x=√9.
Square roots always have two answers, positive and negative, so x=±3.
That means that your solutions are -2, -3 and 3, which agrees with ganesh's workings.
Why did the vector cross the road?
It wanted to be normal.
2sinx.cosx + cos^2x = 1