Math help thanks?

Question

Could you show me the stepsto solve these problems for quadratic inequalities.

1. (x-2)(x-3)(x-4) >0

(x+1)(x+2)(x+3) _< 0

_< is the arrow on top of the line

Answer 1

1.

(x - 2) (x - 3) (x - 4) > 0

To solve this inequality, all you have to do is solve the corresponding equation (x - 2) (x - 3) (x - 4) = 0 to get your critical numbers. In this case, your critical numbers would be
x = {2, 3, 4}.

Now that you have your critical numbers, you want to test a single value in the regions AROUND them, because a single value determines the behavior in that region. In your case, you want to test:
a) a value less than 2
b) a value between 2 and 3
c) a value between 3 and 4
d) a value greater than 4

You're going to plug in this value you're testing in
(x - 2)(x - 3)(x - 4) and determine if it is *positive* (since the right hand side of the inequality states > 0).

Test a value less than 2; 1. Then we have
(1 - 2)(1 - 3)(1 - 4) = (negative) (negative) (negative) = negative.
Therefore, we reject the region (-infinity, 2).

Test a value between 2 and 3: 5/2.
(5/2 - 2) (5/2 - 3) (5/2 - 4) = (positive) (negative) (negative) = positive.
Therefore, include the region (2, 3)

Test a value between 3 and 4; 7/2
(7/2 - 2) (7/2 - 3) (7/2 - 4) = (positive) (positive) (negative) =
negative.
Reject the region (3, 4)

Test a value greater than 4. Let's go extreme and test 1000000.
(1000000 - 2) (1000000 - 3) (1000000 - 4) = definitely positive.
Include the region (4, infinity)

Therefore, your solution set would be everywhere which it is positive. That is,

(2, 3) U (4, infinity)

As a key note to *why* I tested 1000000, the reason is because it doesn't matter what number you test, as long as it fits in the region. Note that by choosing one million, it is VERY easy to tell if your result is positive or negative.

2) (x + 1) (x + 2) (x + 3) <= 0

In this case, we do the same thing, but with one difference; since the inequality states "less than _or equal to_ 0", we're going to *include* the critical numbers in our intevals. Note in the earlier question, we used round brackets; the reason why we used round brackets was because the inequality stated strictly _greater than_ 0.

Critical numbers: x = {-3, -2, -1}

For (x + 1) (x + 2) (x + 3):

Test x = -1000.
Then, we have (-1000 + 1) (-1000 + 2) (-1000 + 3) =
(negative) (negative) (negative) = negative.
Include the region (-infinity, -3]
{NOTE THE SQUARE BRACKET WITH THE -3!}

Test x = -2.5 (since this is between -3 and -2).
Then we have (-2.5 + 1) (-2.5 + 2) (-2.5 + 3) =
(negative) (negative) (positive) = positive.
Reject the region [-3, -2].

Test x = -1.5 (since this is between -2 and -1).
Then we have (-1.5 + 1) (-1.5 + 2) (-1.5 + 3) =
(negative) (positive) (positive) = negative.
Include the region [-2, -1]

Test x = 10000 (since this is greater than 1).
Then we have (10000 + 1) (10000 + 2) (10000 + 3) =
there's no doubt about it that this is POSITIVE.
Reject the region [-1, infinity)

Therefore, our solution set is
(-infinity, -3] U [-2, -1]

****
What you'll learn from experience is that, as long as your factors are of an odd power (in your case, they all had a power of 1), your regions will alternate signs. In your case, note that the regions surrounding (x - 2) (x - 3) (x - 4) shifted from + to - to + to -. What that means is that you really only have to test one region, and then the pattern would emerge.

As something more advance, consider this inequality:

x^2 (x + 5) (x - 5) >= 0

In this case, you have a critical number which is squared, or of an even power. Our critical numbers are x = {-5, 0, 5}.
In this case, what's going to happen is that BOTH regions surrounding the "double root" are going to be positive or negative, but the alternating pattern in the other regions will remain. Just to show you what I mean, let's test a region.

Test -1000. Then we have x^2 (x + 5) (x - 5) being
(positive) (negative) (negative) = positive.

Therefore, it's positive in the region (-infinity, -5]. As per the pattern:
Negative in [-5, 0]
Negative in [0, 5]
Positive in [5, infinity)

Note that before and after 0, it's the same and that it *doesn't* alternate. This is the effect of the "double critical number".

Now that you're armed with this fact, you can solve inequalities like

(x - 1)^2 (x + 2)^3 (x^4) > 0

With the rule of thumb being:
Critical numbers of an odd power alternate; even powers do not alternate.

You'll encounter fractional ones too, but I won't get into those in this answer.

Answer 2

Could you show me the stepsto solve these problems for quadratic inequalities.

First off, these are not quadratic ineqaulities. They are cubic inequalities.

1. (x-2)(x-3)(x-4) >0
The inequality holds for all x >4 sinc all three factors will be >0
If 2 < x < 3 the inequality also holds because the first factor will be >0 and the other two factors will be < 0 so their product is >0.

(x+1)(x+2)(x+3) =< 0
The inequality holds for all x <= -3 and for -2 <= x <= -1
Use the same reasoning as in the first problem.

Answer 3

sure. Multiply out the factors
Graph the result as if it were an equation, not an inequality.

< means everything "under the graph"; > means everthing "over the graph"
the arrow on top of the line means include the actual curve. Without it, then the curve is represented by a dotted line and is not included.

Answer 4

If you meant 2x/(3 - 1) = 7/x 2x/(3 - 1) = 7/x 2x×x = 7×2 2x^2 = 14 x^2 = 7 x = ±√7 so x = 2.64575 or -2.64575 Or did u mean 2x/3 - 1 = 7/x 2x/3 - 1 = 7/x x(2x/3 - 1) = 7 2x^2/3 - x = 7 x^2 - 3x/2 = 21/2 (x - 3/4)^2 = 21/2 + 9/16 (x - 3/4)^2 = 168/16 + 9/16 (x - 3/4)^2 = 17716 x - 3/4 = ±√(177/16) x = 3/4 ± √(177/16) so x = 4.07603 or -2.57603 Hope this helps you. *βω*