Solving inequality questions.?

Question

a) x^2< 2x+8

b) 3-x

------ < 5

x-2

c) |x^5-3| (x^2-x-6)≤ 0

-----------------------

x^2-25

details and tips would be much appreciated!

Answer 1

a)  x^2 &lt; 2x + 8

=&gt; x^2 - 2x - 8 &lt; 0

=&gt; (x - 4)(x + 2) &lt; 0

Therefore x - 4 &lt; 0 =&gt; x &lt; 4

or x + 2 &lt; 0 =&gt; x &lt; -2

Comparing the two solutions we have the solution set as {x | x &lt; 4} since all values less than -2 are also less than 4.  Alternatively, it can be written as &quot;x is an element of [infinity, 4)&quot; since x cannot equal 4, or {x | x &lt; -2}.

b) (3 - x)/(x - 2) &lt; 5

=&gt; 3 - x &lt; 5(x - 2)     noting that x cannot equal 2 for later

=&gt; 3 - x &lt; 5x - 10

=&gt; 6x - 13 &gt; 0

=&gt; 6x &gt; 13

=&gt; x &gt; 13/6

Therefore &quot;x is an element of the interval (13/6,infinity]&quot; or {x | x &gt; 13/6}

c) |x^5 - 3|(x^2 - x - 6)/(x^2 - 25) ≤ 0

=&gt; |x^5 - 3|(x - 3)(x + 2)/[(x + 5)(x - 5)  Ã¢Â‰Â¤ 0

Remove the denominator and note that x = 5 and x = -5 are points of discontinuity (points where the value of x is undefined) {ie, x can&#039;t equal 5 or -5}

|x^5 - 3|(x - 3)(x + 2)  Ã¢Â‰Â¤ 0

|x^5 - 3| will NEVER be negative, so the inequality will only be less than 0 if x - 3 is negative AND x + 2 is positive, or vice versa.

x - 3 is negative when x &lt; 3

x + 2 is negative when x &lt; -2

Both are negative when x &lt; -2 which would make (x - 3)(x + 2) &gt; 0.

Both are positive when x &gt; 3 which would also make (x - 3)(x + 2) &gt; 0.

When x &lt; -2, |x^5 - 3| is not equal to zero either, so x &lt; -2 can NOT be a solution to the inequality.

When x &gt; 3, |x^5 - 3| is not equal to zero either, so x &gt; 3 can NOT be a solution to the inequality either.

When x = fifth root of 3, |x^5 - 3| = 0.

For every other value in the interval [-2, 3], |x^5 - 3| is positive,

whether x = -2 or x = 3, or anything else in between.

Therefore the only possible solutions will occur when x is an element of the interval [-2, 3], or {x | -2 ≤ x ≤ 3}

NOTE - The earlier discovered points of discontinuity can now be ignored because they are not within the solution interval anyway.  There are no points of discontinuity.

Answer 2

The trick is to isolate the x and whatever you do to one side of the inequality, you do to the other.  And remember, if you divide by a negative number, you change the sign.

a) x^2&lt; 2x+8

x^2 -2x -8&lt;0

x-4(x+2)&lt;0

x&lt; 4 or x &lt;-2

b) is this one 3-x/x-2 &lt; 5?  if so, multiply by x-2

3-x&lt; 5(x-2)

3-x&lt;5x -10  add 10 and x to both sides

13 &lt; 6x  divide by 6

13/6 &lt; x or x&gt; 2 1/6