Solve inequality?

Question

answer what you know i got most of it done but got stuck on these?
X^2 - 6X+9 < 16

X^2 +2X > 3

X^2-4x-1 > 0

-2x^2+6X+15 = < 0

X^3+2x^2-4x-8 =< 0

2x^3+13x^2-8x-46 >= 6

4X^3-12^2 > 0

2X^3-x^4 =<0

how do you graph it

Answer 1

Let us solve the 6th inequality which seems to be the hardest one.
2 x³ + 13 x² - 8 x – 46 ≥ 6
x² (2 + 13) - 8 x – 46 ≥ 6
x² (2 + 13) - 8 x – 52 ≥ 0
x² (2 + 13) - 4 ( 2 x + 13) ≥ 0
(x² - 4) (2x + 13) ≥ 0
(x + 2) (x - 2) (2x + 13) ≥ 0
(x + 2) (x - 2) (x + 6.5) ≥ 0
(x + 6.5) (x + 2) (x - 2) ≥ 0

For
x = -6.5
x = -2
and
x = 2
(x + 2) (x - 2) (x + 6.5) = 0
and therefore these three numbers are solutions of the inequality.

Now let us investigate the terms
(x + 6.5),
(x + 2)
and
(x - 2).

For x < -6.5 all three terms are negative and therefore this interval IS NOT a solution of the inequality.
For -6.5 < x < -2 two terms of three are negative and therefore this interval IS a solution of the inequality.
For -2 < x < 2 one term only is negative and therefore this interval IS NOT a solution of the inequality.
For 2 < x all three terms are positive and therefore this interval IS a solution of the inequality.
-
Conclusion: The solution of the inequality are the following two intervals
-6.5 ≤ x ≤ -2
and
x ≥ 2.
-

Answer 2

1.
x^2 -6x -7 < 0
(x-7)(x+1)<0
Critical values: x = 7, -1
0^2 -6(0) -7 < 0
inequality is true for x = 0 (a value between 7 and -1
-2^2 -6(-2) -7 is not less than 0
inequality is false for x = -2 (a value less than -1)
8^2 -6(8) -7 is not less than 0
inequality is false for x = 8 (a value greater than 7)
Answer:
-1
2.
x^2 + 2x -3 > 0
(x+3)(x-1) > 0
Critical values: x =-3, 1
Do similar procedure as in 1.
Answer:
-3 > x > 1

3.
x^2 -4x -1 > 0
You can't factor, so use Pythagorean theorem to get critical values.
x = 2 +/- sqrt(5)
Do similar procedure as in 1.
Answer:
x < 2 - sqrt(5)
x > 2 + sqrt(5)

4.
-2x^2 + 6x + 15 =< 0
Use Pythagorean Theorem to get critical values:
x = 1.5 +/- 0.5 sqrt(39)
Do similar procedure as 1.
Answer:
x >= 1.5 + 0.5 sqrt(39)
x <= 1.5 - 0.5 sqrt(39)

5.
x^3 +2x^2 -4x -8 <= 0
(x+2)(x^2 - 4 ) <= 0
critical values x = +/- 2
Do similar procedure as 1
Answer:
x <= 2

6.
2x^3 +13x^2 -8x-52 >= 0
(2x+13)(x^2-4) >= 0
Critical values: x = -13/2, 2, -2
Do similar procedure as in 1.
Answer:
x >= 2

7.
4x^3 -12x^2 > 0
2x^3 -6x^2 > 0
2x^2 (x - 3) > 0
Critical values: x = 0, 3
Do similar procedure as 1.
Answer:
x > 3

8.
x^3 (2-x) <= 0
Critical values: x = 0, 2
Try x = -1 (a value less than 0)
-1 (2+ 1) = -3 <= 0
Inequality is true for x <= 0
Try x = 1 (a value betweeen 0 and 2)
1 (2 - 1) = 1 is not less than or equal to 0.
Inequality is false between 0 and 2.
Try x = 3 (a value greater than 2)
81 (2 - 3) = -27 <= 0
Inequality is true for x >= 2.
Answer:
x <= 0, x >= 2

If you put zero on one side of the inequalities and treat the other side as a function, you can graph it and you will see that the the solution to the inequality is all values of x for which the graph is above or below the x axis.

How do you graph these functions? If you are taking calculus, you can take the derivative, set it equal to zero and find the intervals over which the function is increaing or decreasing. You can also plot it on a graphing calculator. If it is a quadratic, you can put it in standard form:
y = ax^2 + bx + c
Your axis of symmetry is x = -b/2a and you can find the vertex by plugging in this value into the equation. If a < 0, the parabola opens downward (looks sort of like a mountain). If a > 0, it opens upward (looks sort of like a valley).

If you have access to a spreadsheet, like Excel, you can set up a table of x's and y's like the one below (#6)
a1: 0.01
a2: x __________ y
a3: 0.00_______ = 2*A3^3+13*A3^2-8*A3-52
a4: = a3+$a$1__ = 2*A4^3+13*A4^2-8*A4-52
a5: = a4+$a$1__ = 2*A5^3+13*A5^2-8*A5-52
a6: = a5+$a$1__ = 2*A6^3+13*A6^2-8*A6-52

Your table will look like this:
dx 0.01
x y
-4.7 65.124
-4.69 65.145882
-4.68 65.164736
-4.67 65.180574
-4.66 65.193408
-4.65 65.20325
-4.64 65.210112

The graph looks like a mountain followed by a valley.

Answer 3

1. x^2-6x+9-16<0
x^2-6x-7<0
(x-7)(x+1)<0
x-7=0, x=7
x+1=0, x=-1.
First look at value <-1.
(-2-7)(-2+1) = -14*-1=14 not < 0.
Now between -1 and 7.
(0-7)(0+1)=-7*1=-7 < 0
Now value > 7.
(8-7)(8+1)=1*9=9 not < 0.
SO, its -1
2.x^2+2x-3>0
(x+3)(x-1)>0
x=-3 , x=1.
Value <-3.
(-5+3)(-5-1) =-2*-6=12 >0.

Value between -3 and 1.
(0+3)(0-1) = 3*-1=-3 not > 0.

value > 1
(2+3)(2-1)=5*1=5>0.

SO, its x<-3 and x>1.

3.-2x^2+6x+15=<0.
Use quadratic formula
x=(-6+/-sqr(6^2-(4*-2*15))) /(2*-2) =
(-6+/-sqr156)/-4 = (-6+-12.48999)/-4.
SO, (-6+12.48999)/-4 = -1.6225
and (-6-12.48999)/-4 = 4.6225.
Do same as above,
with value < -1.6225
value between -1.6225 and 4.6225
and value > 4.6225
So, its x<-1.6225 and x> 4.6225

4.x^3+2x^2-4x-8=<0
(x-2)(x+2)^2=<0
SO, x=2 and x=-2.
Value between -2 and 2
0+0-0-8=<0.
SO, its -2<=x<=2

5.2x^3+13x^2-8x-46-6>=0
2x^3+13x^2-8x-52>=0
(x-2)(x+2)(2x+13)>=0
So, x=2, x=-2, x=-13/2.
SO, test value between -2 and 2.
0+0-0-52 not >= 0.
SO, I think?
-13/2<=x<=-2
and x=>2.

6.4(x^3-36)>0

7.x^3(2-x)=<0
x=0, x=2.
And do as above.
I think?
x<=0
x>=2.

Answer 4

you need to use the quadratic equation. for instance on the first one: subtract 16 on both sides to get: x^2 - 6x -5 < 0
then use the quadratic equation to get x