Help on this problem!! x^3 + 3x < 4x^2?

Question

Directions: Solve the inequality in terms of intervals and illustrate the sollution set on the real number line.

1.) x^3 + 3x < 4x^2

2.) x^2 < 2x + 8

3.) 2x - 3 < x + 4 < 3x - 2

4.) 1 + 5x > 5 - 3x

Answer 1

Hi,

1.) x³ + 3x &lt; 4x²

x³ - 4x² + 3x &lt; 0

x(x² - 4x + 3) &lt; 0

x(x - 3)(x - 1) &lt; 0

x intercepts are at 0, 1, and 3.

Since a cubic inequality with a positive leading coefficient comes up from negative infinity on the left and shoots up to positive infinity on the right side, the graph is located as follows:

....... .0 above 1.... ....2..... .....3 above

--------+------ ----+--------+-----.-----+----- -----

below ....... ..... below . below

........._____....... ....... ....... ..../

......0/ .... ...\1.... ....2..... .....3/

-----+------ ----+--------+-----.-----+----- -----

..../....... .......\....... ....... ../

../....... ....... ...\________/

/

The solution set is (x | x &lt; 0 or 1 &lt; x &lt; 3). &lt;==ANSWER

2.) x² &lt; 2x + 8

x² - 2x - 8 &lt; 0

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

x intercepts are x = -2 and x = 4.

Since this parabolic equation opens up, the only part below the x axis is between the intercepts.

The solution set is (x |  -2 &lt; x &lt; 4). &lt;==ANSWER

3.) 2x - 3 &lt; x + 4 &lt; 3x - 2

2x - 3 &lt; x + 4 and x + 4 &lt; 3x - 2

x &lt; 7 and x &gt; 3

The solution set is (x |  3 &lt; x &lt; 7). &lt;==ANSWER

4.) 1 + 5x &gt; 5 - 3x

1 + 8x &gt; 5

8x &gt; 4

x &gt; ½

The solution set is (x | x &gt; ½). &lt;==ANSWER

I hope that helps!! :-)

Answer 2

what you do with these types of problems is to first solve it as if the &quot;greater/less than&quot; were the same as &quot;equal&quot; signs.

For example:

1&gt;   x^3+3x&lt;4x^2

rewrite them as:    x^3+3x=4x^2

solve:                  x^3-4x^2+3x=0

                          x(x^2-4x+3)=0

                          x(x-3)(x-1)=0

                           x = 0, 3, 1

now you got the 3 points.  these 3 points indicates the end points of 4 sections on a number line.

So the possible answers are from   negative infinity to 0,  0 to 1,  1 to 3, and 3 to infinity

so test numbers that lies within those regions to see if it still makes your original question true.

For &quot;negative infinity to 0&quot;, let&#039;s pick -10 (for ease of calculation)

Is (-10)^3+3(-10) &lt;? 4(-10)^2    ? in your original equaiton?

   -1000-30 &lt;?400               -1030 &lt; 400     True  (Part of the answer)

let&#039;s test next section:  0 to 1

let&#039;s pick  0.1 (for ease of calculation)

(0.1)^3 + 3(0.1) &lt;? 4(0.1)^2   ?

0.001 +3 &lt;?  4*0.01                     3.001&lt;? 0.04       False

So you know this region is not part of the answer.

next region:  

1 to 3:   let&#039;s pick 2

2^3+3(2)&lt;?4(2^2)                8+6&lt;?16           14&lt;16     True

next region:

3 to infinity:  let&#039;s pick 10   (for ease of calculation)

10^3 +3(10) &lt;?  4(10^2)

1000+30 &lt;? 400            1030 &lt;? 400       False

so we know that only the 1st and 3rd group of our answers are true.

Thus our answers are regions from negative infinity to 0,  and 1 to 3.

good luck