# Thread:

1. ## inequalities

Please check my work.thank you so much
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2. Quadratic Inequalities are always easiest to solve if you complete the square.

$\displaystyle x^2 - 2x - 3 \leq 0$

$\displaystyle x^2 - 2x + (-1)^2 - (-1)^2 - 3 \leq 0$

$\displaystyle (x - 1)^2 - 4 \leq 0$

$\displaystyle (x - 1)^2 \leq 4$

$\displaystyle \sqrt{(x-1)^2} \leq \sqrt{4}$

$\displaystyle |x-1| \leq 2$

$\displaystyle -2 \leq x - 1 \leq 2$

$\displaystyle -1 \leq x \leq 3$.

3. Another good way to solve quadratic inequalities (and other complicated inequalities) is to solve the associated equation.

For example, to solve $\displaystyle x^2 - 2x - 3 \leq 0$, first solve $\displaystyle x^2 - 2x - 3= 0$, by factoring, $(x- 3)(x+ 1)= 0$ 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 $(-2)^2- 2(-2)- 3= 4+ 4- 3= 5> 0$ so the polynomial is greater than 0 for all x< -2. x= 0 is between -1 and 3 and $(0)^2- 2(0)- 3=-3< 0$ so the polynomial is less than 0 for all x between -1 and 3. Finally, x= 4 is larger than 3 and $(4)^2- 2(4)- 3= 5> 0$ so the polynomial is greater than 0 for all x greater than 3. Since the inequality say " $\le 0$", the solution set is $-1\le x\le 3$.

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.

4. Originally Posted by jam2011
Please check my work.thank you so much
$x^2-2x-3\ \le\ 0$

$(x-3)(x+1)\le\ 0$

If $x-3 \ge\ 0$, then $x+1=(x-3)+4$ must be $\ge\ 0$ since it is 4 greater than (x-3)

No luck there...

Case 2:

If $x-3\ \le\ 0$, then $x\ \le\ 3$ and $x+1\ \ge\ 0$ only for $x\ \ge\ -1$

Hence $-1\ \le\ x\ \le\ 3$

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

Yet another way to proceed is....

$x^2-2x\ \le\ 3\Rightarrow\ x(x-2)\ \le\ 3$

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

$x(x-2)=3$ for $x=3$ and for $x=-1$

If $x<-1$, then $x(x-2)>3$

If $x>3$, then $x(x-2)>3$

The solution is $-1\ \le\ x\ \le\ 3$

5. $x^2 - 2x - 3 \le 0$

$(x + 1)(x - 3) \le 0$

think about the graph of the quadratic ... a parabola that opens upward with two zeros, $x = -1$ and $x = 3$

the section of the graph below the x-axis (the less than or equal to section) lies between and includes the two zeros.

6. 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...
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