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 Apr 26, 2020
edited by VVoidSmiley69  May 27, 2020
 #1
avatar+21017 
+2

Setting the denominator to zero will break up the domain into 5 parts.

2x2 + x - 6  =  0     --->     (2x - 3)(x + 2)  =  0     --->     x = 1.5  and  x = -2.

The five parts are:  1)  the region below x = -2

                                2)  x = -2                                                              <--- at this point there is a break in the graph

                                3)  the region between x = -2 and x = 1.5

                                4)  x = 1.5                                                            <--- at this point there is a break in the graph

                                5)  the region above x = 1.5

 

Choose a number in the region below x = -2  ---   any number, I chose -5.  Place this number into the problem to see

if the answer is above 0 or below 0. It's above 0, so this region "works".

 

Choose a number in the region between x = -2 and x = 1.5  ---   any number, I chose 0. Place this number into the

problem to see if the answer is above 0 or below 0. It's below 0, so this region "doesn't work".

 

Choose a number in the region above x = 1.5  ---   any number, I chose 5.  Place this number into the problem to see

if the answer is above 0 or below 0. It's above 0, so this region "works".

 

So, the answer is:  (-infinity, -2)  union  (1.5, infinity)

 Apr 26, 2020
 #2
avatar+84 
+2

Thanks a lot Geno, I clearly understand the problem now. :)

VVoidSmiley69  Apr 26, 2020
 #3
avatar+109527 
+2

Here is my way, it is similar, just a bit different

 

\(\frac{x^2 + x + 3}{2x^2 + x - 6} \ge 0. \)

 

 

Think of the numerator and denominator separately. 

They both have to be positive OR they both have to be negative in order for the final answer to be positive.

 

If you let  y=x^2+x+3 

first notice it is a parabola.

Solve it for y=0 you will find there are no real answers, the discriminant under the square root sign is negative.

Since the leading coefficient is + it must be a concave up.

A concave up parabola with no real roots MUST always have a positive y value.

So I have discovered that the numerator (top) is always positive.

 

So the bottom must be positive too.

the denominator (bottom) is also a concave up parabola when set equal to y.

If there are any roots then the positive bit must be on either end of the roots (not in the middle)

Solving for 0=2x^2+x+6 gives  x=-2 and x=1.5

 

SO  the expression will be positive so for all values of x less than -2 or more than 1.5

 Apr 26, 2020
 #4
avatar+84 
+2

Thanks Melody, this answer als provides me a different insight the problem. cool

VVoidSmiley69  Apr 27, 2020

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