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# Find the solution set of the given equation parabola

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1.) x3+27/x2-3x-10 < 0
2.) (3x-1) (2-5x)/2x+1 > 0
3.) 2/x+2< or equal to x/x+1

Guest Jul 16, 2017
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#1
+83935
+1

I'm assuming this is

1.)   (x^3+27) / (x^2-3x-10) < 0     factoring the numerator and denominator, we have

[(x + 3) (x^2  - 3x + 9)] / [ ( x - 5) (x + 2) ] < 0

The second factor in the numerator is always positive so the numerator is < 0 whenever x < -3

The denominator is undefined whenever x = 5 or x = -2

And the denominator will < 0  whenever    -2 < x < 5

So......the function will be < 0 whenever the numerator is positive and the denominator is negative  or when  the numerator is negative and the denominator is positive

The first situation will occur when    -2 < x < 5

And the second wil occur when    x < - 3

So the solution set is  [  x < - 3 ]    U   [    -2 < x < 5  ]

Here's the graph :

https://www.desmos.com/calculator/xgh0b1qmvm

CPhill  Jul 16, 2017
#2
+83935
+1

I'm assuming that we have :

2.)   (3x-1) (2-5x) / (2x+1) > 0

Setting each factor to 0 and solving for x, the critical values are :

x = 1/3 , x = 2/5  and x = -1/2

The function will be undefined when x = - 1 /2

The numerator is positive when    1/3 < x < 2/5

And the denominator is positive when x > -1/2

And the function is > 0 whenever the numerator and denominator have the same sign

So...the function is > 0  when

1/3 < x < 2/5       or when     x < -1/2

Here's the graph :

https://www.desmos.com/calculator/utup8jcqp2

CPhill  Jul 16, 2017
#3
+83935
+1

I'm assuming that the last one is :

2 / (x+2)   ≤   x / (x+1)

Setting each factor to 0 and solving for x, the critical values are  -2  and - 1

Note that the inequality is undefined at these two x values

Turn this into an equality

Cross-multiplying, we have

2 (x + 1)  =  x ( x + 2)     simplify

2x + 2  =  x^2 + 2x     subtract 2x from both sides

2  = x^2     rearranging, we have

x^2 - 2 = 0

And this will be true when   x = - √2   or   x = √2

And these are the other two critical values that we need

Note that the inequality will  be true when

x < -2 ,     - √2   ≤ x < -1   and  x ≥ √2

Here's a graph :

https://www.desmos.com/calculator/zu0vmqfgux

CPhill  Jul 16, 2017

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