Please check my work.thank you so much

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- Jan 31st 2011, 10:16 PMjam2011inequalities
Please check my work.thank you so much

- Jan 31st 2011, 10:27 PMProve It
Quadratic Inequalities are always easiest to solve if you complete the square.

. - Feb 1st 2011, 11:47 AMHallsofIvy
Another good way to solve quadratic inequalities (and other complicated inequalities) is to solve the associated

**equation**.

For example, to solve , first solve , by factoring, so x= 3 and x= -1. The point is this: a polynomial is a "continuous" function which means that it in order to go from positive to negative, it**must**pass through 0 and that can happen only at x= -1 and x= 3 which divide the real line into three intervals.

To decide which interval is ">0" and which "< 0" you can use either of two methods:

1) Choose a number in each interval and check the sign. x= -2 is less than -1 and so the polynomial is greater than 0 for**all**x< -2. x= 0 is between -1 and 3 and so the polynomial is less than 0 for all x between -1 and 3. Finally, x= 4 is larger than 3 and so the polynomial is greater than 0 for all x greater than 3. Since the inequality say " ", the solution set is .

2) Notice that x+ 1 is negative for x< -1 and positive for x> -1 and that x- 3 is negative for x< 3 and positive for x> 3. If x< -1, it is also less than 3 so both factors are negative and their product is positive. If -1< x< 3, one factor, x+1, is positive but x- 3 is still negative. The product of one positive and one negative factor is negative. Finally, if x> 3, both factors are positive and their product is positive. That gives the same solution set as before and (fortunately!) the same solution as Prove It. - Feb 1st 2011, 01:38 PMArchie Meade

Your Case 1:

If , then must be since it is 4 greater than (x-3)

No luck there...

Case 2:

If , then and only for

Hence

so your answer is correct and the diagrams help illustrate the intersections.

**Yet another way to proceed is....**

The factors of 3 that differ by 2 are 1 and 3 or and

for and for

If , then

If , then

The solution is - Feb 1st 2011, 03:03 PMskeeter

think about the graph of the quadratic ... a parabola that opens upward with two zeros, and

the section of the graph below the x-axis (the less than or equal to section) lies between and includes the two zeros. - Feb 1st 2011, 03:08 PMjam2011
nice job !!! thank you so much...i learn a lot.there are many ways to find and solve this problem. i owe you a lot...