math question?!?!?

Question

X^3 + 3X < 4(X^2)
it's read --
Xcubed plus three X is less than four X squared.

how would you solve?

Answer 1

begin by subtracting 4(X^2) from both sides and you get
X^3 + 3X - 4(X^2) < 0
which is the same as :
X(X^2 + 3 -4X) <0
X^2 + 3 -4X can be factorized as (X-3)(X-1)
(you can get a deeper explanation of that particular factorization at http://en.wikipedia.org/wiki/Factorization)
So now you have to solve:
X(X-3)(X-1) <0
if X>3, all three factors are positive and the inequality is wrong
if 1 if 0 if X<0, all three factors are negative and the inequality is right.
So, your inequality is right when :
X<0 and 1

Answer 2

x^3 + 3x < 4x^2

First set the inequality less than 0 by moving all the terms to one side. Subtract 4x^2 from both sides.

x ^3 +3x - 4x^2 < 0.

Now you have a polynomial inequality of degree 3, which means that you will have 3 solutions. The best way to solve now would be to look for a common factor among the three terms, which in this case will be x. After you factor out an x, the inequality will be

x (x^2 - 4x + 3) < 0. (Notice the terms are rearranged in order of greatest to least degree.)

Now factor the remaining quadratic expression by looking for factors of 3 that add up to -4. After factoring, the inequality will be

(x) (x-1) (x-3) < 0

Using the zero product property, set each term less than zero and solve for x.

x < 0

x - 1< 0
x < 1

x - 3 < 0
x < 3

If this were an equation rather than an inequality, you would be finished at this point. However, we need to figure out for which values of x that the product (x) (x-1) (x-3) is less than zero. The easiest way that I have found to do this is to construct a chart to show for what values each term is positive or negative. (I couldn't get this typeset correctly; the 0 is supposed to be above the place where the signs change in the first row, the 1 above the sign change in the second row, and the 3 above the sign change in the third row.)

0 1 3

x -----+++++++++++++++++
x-1 ---------------++++++++++
x-3 -------------------------++++

Now notice that the chart can be divided into four regions (from negative infinity to 0, from 0 to 1, from 1 to 3, and from 3 to positive infinity). From negative infinity to 0, the answer will be negative (three negative numbers multiplied together). Negative numbers are less than zero, so that region satisfies our original inequality. The region from 0 to 1 does not (negative * negative * positive = positive), the region from 1 to 3 does (positive * positive * negative = negative), and the region from 3 to positive infinity does not (positive * positive * positive = positive). The answer will contain the two regions that did satisfy the inequality written in interval notation.

(negative infinity, 0), (1,3)

Hope this helps.

Answer 3

First, shift 4(x^2) to the right side of the inequality.
x^3 + 3x - 4(x^2) < 0
Now, factor out an 'x'
x(x^2 - 4x + 3) < 0
Now you have two inequalities:
x < 0 AND x^2 -4x + 3 < 0
The second inequality can be solved with the quadratic formula:
(x - 1/3)(x + 4/3) < 0
Which gives two new inequalities:
x < 1/3 and x < -4/3
So, we have the following inequalities for x:
x < 1/3 , x < 0 , x < -4/3

Answer 4

x^3 + 3x < 4x^2

subtract 4x^2 for both sides
x^3 - 4x^2 + 3x < 0

factor out x
x (x^2 - 4x + 3) < 0

factor
x (x - 3) (x - 1) < 0

x = 0, 1, 3

now you need to plug some numbers in to see which intervals make the whole equation less than 0

0 ................................. 1 ................................... 3

plug in -1
-1 (-1 - 3) (-1 - 1) -1(-4) (-2) -8 < 0

the statement is true. Thus x < 0

now plug .5
.5 (.5 - 3) (.5 - 1) .5 (-2.5) (-.5) .625 < 0

the statement is false.

Now plug 2
2(2 - 3) (2 - 1) 2(-1) (1) -2 < 0

the state meant is true. thus, x > 1

now plug in 4

4 (4 - 3) (4 - 1) 4(1)(2) 8 < 0

the statement is false, thus x < 3.


the solutions are: x < 0 and 1 < x < 3

Answer 5

x^3-4x^2+3x < 0
x(x^2-4x+3) < 0
x(x-3)(x-1) < 0
basically, you looking at critical points of 0, 3, 1; pick test points between these intervals such as -1, .5, 2, 4 and plug in

x=-1, you get a negative #
x=.5, you get a positive #
x=2, you get a negative #
x=4, you get a positive #

You want the negative # b/c that's what is < 0. So, the answer is...

(-infinity, 0) U (1, 3)

Answer 6

x^3 + 3x < 4(x^2)
x(x^2 + 3) < x(4x) divide out an x
x^2 + 3 < 4x move the 4x to the other side
x^2 - 4x +3 < 0 factor
(x-3)(x-1) < 0

Answer 7

X^3 + 3X < 4(X^2)
X^3 -4X^2 +3X < 0
X(X^2-4X+3)<0
X(X-1)(X-3) < 0
x<0 1

Answer 8

x^3+3x<4x^2
Divide by x
x^2+3<4x
Subtract 4x
x^2-4x+3<0
Factor the quadratic equation
(x-3)(x-1)<0
To be less than 0, you must multiply by a (-)number
x<1
x<3