Solving inequalities ?

Question

answer in back says (-2/3)<=x<=(1/2) question is (2x-1)(3x+2)<=0 i got..x<=(1/2) and x<=(2/3) did i do something wrong the way i did it was i did each bracket seperately so 2-1<=0 is x<=(1/2) and 3x+2<=0 is x<=(-2/3) so why did i get it wrong

Answer 1

You don&#039;t want to handle each factor separately

(at least not in the limited fashion you&#039;ve used).

Take each factor and sketch its behavior over a numer line:

- - - - - - - - - - -(1/2)++++++++++ for (2x-1)

- - - - (-2/3)+++++++++++++++++ for (3x+2)

+++++(0)- - - - - (0)+++++++++++ for product

Then notice the sign you&#039;d get in each of the 3 apparent intervals from multiplying the factors (2 in your case):

+ in the interval left of (-2/3); zero or negative in the closed interval [-2/3, 1/2] ; then positive right of 1/2.

Since you want the product to be zero or negative you choose the interval that fits your requirements:  [-2/3, 1/2]

Answer 2

When x is less than -2/3, (and therefor less than 1/2) you have two negatives.  

The product of 2 negs is greater than 0.

so x must be greater than -2/3 for the statment to be true.

What you did was to find the places where the 2 functions individually satisfy the &lt;0.  But those 2 functions interact with each other because of their signs.

Answer 3

i got the same thing you got. you know sometimes the book is wrong...

Answer 4

you shouldn&#039;t calculate it separately. that&#039;s wrong way.

1) assume (2x-1)(3x+2) = 0 to find the critical point(s).

2) sketch the parabola of the equation.

3) determine the region of x: equation&lt;=0, the interval x is between the parabola: equation&gt;=0, the interval x is outside the parabola

you already find the critical x values which are 1/2 and -2/3.

since the equation is &lt;=0, so the interval x should between 1/2 and -2/3. logically, you will find that -2/3&lt;=x&lt;=1/2.