Inequality problem?

Question

can someone help me solve this inequality. I did it, but I dont think I did it right?

x^2 - 6x -2 < x - 8

thanks!

Answer 1

I think everyone has given you the method of getting to the factorization... basically just move the x - 8 to the other side (subtract x and add 8).

That gets you to:
x² - 7x + 6 < 0

This factors as:
(x - 6)(x - 1) < 0

However there have been several conflicting answers on where to go from here. Notice that you will have a negative product when exactly one of the expressions in the parentheses is negative. But if they are *both* negative, or *both* positive, you will have the product being greater than 0.

You have two possible cases:

CASE 1:

(x - 6) > 0 and (x - 1) < 0

The solution for this would be:
x > 6 and x < 1

But there is no way for x to be greater than 6 and less than 1. You can throw out this case.

CASE 2:

(x - 6) < 0 and (x - 1) > 0

Solution for this is:
x < 6 and x > 1

Rewriting you get:
1 < x < 6

This is the correct solution, and I think one other answerer had it.

Edit: I see that kindricko was replying at the same time I was and also has the correct answer.

You can also double-check the possible answers. Notice that if you try a value of x less than 1 (say x = 0), you end up with -2 < -8, which is not correct. Or if you try a value > 6 (say x = 7) you end up with 5 < -1, which again is incorrect.

The correct answer is x is between 1 and 6 (but not equal).
1 < x < 6

Answer 2

x^2 - 6x - 2 < x - 8

x^2 - 7x - 2 < - 8

x^2 - 7x + 6 < 0

(x - 1)(x - 6) < 0

In order for this to be true, one factor must be positive and the other factor must be negative

Case 1

x - 1 > 0
x > 1

x - 6 < 0
x < 6

x can be > 1 and < 6

1 < x < 6


Case 2

x - 1 < 0
x < 1

x - 6 > 0
x > 6

x can't be < 1 and > 6 at the same time

The case 1 range is the best answer, I think

Answer 3

x^2 - 6x - 2 < x - 8
x^2 - 6x - x - 2 + 8 < 0
x^2 - 7x + 6 < 0

Factorise this quadratic expression using mid term breaking:

x^2 - 6x - x + 6 < 0
x (x - 6) - (x - 6) < 0
(x - 1)(x - 6) < 0
x - 1 < 0 OR x - 6 < 0
x < 1 OR x < 6

x < 6 implies that all the values x < 1 also come into this interval because all the values less than 1 are also less than 6.

Hence the answer is: x < 6

Answer 4

x^2 - 6x -2 < x - 8
x^2 - 7x + 6 < 0
(x-1)(x-6)<0

--> 1

Answer 5

x^2-7x-10<0


*since this is horrible to factor, i'm using the quadratic equation*
x=(-b+/-sqrt(b^2-4ac))/2a

x=(-(-7)+/-sqrt((-7)^2-4(1)(-10))/2(1)

x=(7+/-sqrt(49+40))/2

x=(7+sqrt(89))/2
x=(7-sqrt(89))/2

*the answers can be turned into factors*
so
x - (7+sqrt(89))/2 = 0
x - (7-sqrt(89))/2 = 0

*rewrite the quadratic inequality using the factors*
[x - (7+sqrt(89))/2][x - (7-sqrt(89))/2]<0

split it
x - (7+sqrt(89))/2 < 0
x - (7-sqrt(89))/2 > 0

*i may have made an error, but it looks to be accurate. one has to be >0 (positive) and the other <0 (negative) for the product to be <0 (negative)
*and subtracting the smaller number should be >0 then and subtracting the larger number should be <0.

Answer 6

BIGGER problem to solve,.HELP,.
http://www.mathematics.com
http://www.mathematics.ca
p.s. your Inequality problem question, the way you are asking may be confusing to some,.I had a hard time understanding your wording, hope these links help, if not e-mail your math teacher,. I do know you are in school,.
---good luck,.. bye,..

Answer 7

subtract x-8 from both sides, leaving you with
x squared -7x+6, then factor it
(x-6)(x-1)<0
x<1,6

Answer 8

the last guy is wrong! digital whatever guy