MHB Solving Polynomial Inequalities

AI Thread Summary
To solve the polynomial inequality $(x - 3)(x + 1) + (x - 3)(x + 2) \ge 0$, first factor the left-hand side to get $(x-3)(2x+3) \geq 0$. Identify the zeros of the function, which are at $x=3$ and $x=-\frac{3}{2}$. Create an interval table to determine where the product $(x-3)(2x+3)$ is positive. The solution will be the intervals where the function is greater than or equal to zero.
eleventhxhour
Messages
73
Reaction score
0
Solve the following inequality:

6e) $(x - 3)(x + 1) + (x - 3)(x + 2) \ge 0$

So, I created an interval table with the zeros x-3, x+1, x-3 and x+2 but I keep getting the wrong answer. Could someone help? (this is grade 12 math - so please don't be too complicated).

Thanks.
 
Mathematics news on Phys.org
I would start with factoring the LHS:
$$(x-3)[(x+1)+(x+2)] \geq 0$$
$$\Leftrightarrow (x-3)(2x+3) \geq 0$$

Now create an interval table and look where the function $f(x)=(x-3)(2x+3)$ which has two zeros at $x=3$ and $x=\frac{-3}{2}$ is positive. That will give you the answer.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.

Similar threads

Replies
13
Views
2K
Replies
5
Views
1K
Replies
7
Views
2K
Replies
2
Views
1K
Back
Top