find the solution set using interval notation. |10 - 4x / x2 - 7x + 10| >3?

Question

|10 - 4x |
| x2 -7x +10 | >3 NOTE!! (its something like this, x2 by the way is suppose to be x square. this by the way is inequalities in calculus)

Answer 1

.. |10 - 4x|
---------------- > 3
|x² - 7x +10|

So

Now |10 - 4x| = 10 - 4x for 10 - 4x ≥ 0 ie x ≤ 2.5
and = -(10 - 4x) for 10 - 4x < 0 ie x > 2.5

Also 3|x² - 7x +10| = 3(x² - 7x +10) for x² - 7x +10 ≥ 0 ie x ≤ 2 and x ≥ 5
and = -3(x² - 7x +10) for x² - 7x +10 < 0 ie 2 < x < 5

Find points of intertsection:

When x < 2
10 - 4x = 3(x² - 7x +10)
3x² - 17x +20 = 0
(x - 4)(3x - 5) = 0
Only solution in domain is x = 5/3

When 2 < x ≤ 2.5
10 - 4x = -3(x² - 7x +10)
So 3x² - 25x + 40 = 0
x = (25 ± √(625 - 480)/6
Only solution in domain is x = (25 - √(145))/6 (≈ 2.160)

When 2.5 ≤ x < 5
4x - 10 = -3(x² - 7x +10)
So 3x² - 17x +20 = 0
(x - 4)(3x - 5) = 0
Only solution in the domain is x = 4

When x > 5
4x - 10 = 3(x² - 7x +10)
So3x² - 25x + 40 = 0
x = (25 ± √(625 - 480)/6
Only solution in domain is x = (25 + √(145))/6 (≈ 6.174)

So
.. |10 - 4x|
---------------- > 3 occurs when |10 - 4x| > 3|x² - 7x +10|
|x² - 7x +10|

and this is when 5/3 < x < (25 - √(145))/6
and when 4 < x < (25 + √(145))/6

WHHHOOOOOOOOPPPSSS YESSSSSSSSSSSSS asymptotes!! at 2 and 5!! so x cannot be 2 or 5

5/3 < x < 2, 2 < x < (25 - √(145))/6, 4 < x < 5, 5 < x < (25 + √(145))/6 (although you can always argue that any number > -∞ and ignore the discontinuities)

Answer 2

|(10 - 4x)/(x^2 - 7x + 10)| > 3
|(-4x + 10)/(x^2 - 7x + 10)| > 3
|-2(2x + 5)/((x - 5)(x - 2))| > 3

since this problem can't be simplified, lets continue

|(-4x + 10)/(x^2 - 7x + 10)| > 3

This can be written as

±(-4x + 10)/(x^2 - 7x + 10) > 3

(-4x + 10)/(x^2 - 7x + 10) > 3
(-4x + 10)/(x^2 - 7x + 10) < -3

-4x + 10 > 3(x^2 - 7x + 10)
-4x + 10 > 3x^2 - 21x + 30
3x^2 - 17x + 20 < 0

x = (-b ± sqrt(b^2 - 4ac))/(2a)
x = (-(-17) ± sqrt((-17)^2 - 4(3)(20)))/(2(3))
x = (17 ± sqrt(289 - 240))/6
x = (17 ± sqrt(49))/6
x = (17 ± 7)/6
x = (10/6) or (24/6)
x = (5/3) or 4

-4x + 10 < -3(x^2 - 7x + 10)
-4x + 10 < -3x^2 + 21x - 30
3x^2 - 25x + 40 < 0

x = (-(-25) ± sqrt((-25)^2 - 4(3)(40)))/(2(3))
x = (25 ± sqrt(625 - 480))/6
x = (25 ± sqrt(145))/6

Now just plug in some values above and below the x values until you come up with a number that doesn't fit, and see what fits the problem.

For example, go from 4 up or 4 down, if one of those works, go in that direction until you run into a number that doesn't fit.

ANS :
(5/3) < x < 2
4 < x < 5
2 < x < ((25 - sqrt(145))/6)
5 < x < ((25 + sqrt(145))/6)

The reason for the answers, is because once you get to 2 or 5, the problem becomes undefined and therefore there are unsolvable gaps at x = 2 and 5.

Answer 3

To have abs val greater than 3, it must be between -3 and +3.

First solve (10 - 4x) /( x^2 - 7x + 10) = 3 and
(10 - 4x) /( x^2 - 7x + 10) = -3

The top one becomes
3x^2 - 21x + 30 = 10 - 4x, and I expect you can solve this to get
x = 5/3 or 4.

Solve the second one and get
???irrational roots?!!! This can't be real! It's possible to do, but horrible!
In fact it's x = (25-sqrt(145))/6 [abt 2.17] or (25-sqrt(145))/6 [abt 6.17]

Another pain is that besides zeros the expression has two asymptotes: The values x = 2 and x = 5 give a zero denominator.

Now evaluate the expression for x = 0, 1.8, 2.1, 3, 4.5, 6, 7, and see whether the absolute value is greater than 3. Why those values? Put the six values I worked out above on the number line and see that each of these values represents a different section of the number line as it's cut up by those six values.
I expect you will find that it's greater than 3 for x = 1.8, 2.1, 4.5, 6, and so the solution is all the bits of the number line containing those values (you could have picked other values as long as they were in those intervals.) i.e. solution
5/3 < x < 2; 2 < x < (25-sqrt(145))/6; 4 < x < 5; 5 < x < (25+sqrt(145))/6

Answer 4

if the x2 is meant to be x^2, then there are 2 ideas. ingredient x^2 + 7x + 10 = 0 to get (x + 5) (x + 2) = 0 set each and each ingredient = 0 --> (x + 5) = 0 --> x = -5 and (x + 2) = 0 --? x = -2 So x = -5 and x = -2 are the two ideas