1. ## interval

1) How do i write the answer in interval notation for (x-5)(x-11)>0 I got (5,infinity)U(11,infinity) webwork said it is incorrect too, so can anyone help me out what mistake i made for those two problems?
2)how will the problem be solved for the inequality in interval notation for -2x^2<18
3)also for the problem -x^2+4x(equal or > to)0 ?

2. Hello, Girlaaaaaaaa!

With quadratic inequalities, we must be careful . . .

$1)\;(x-5)(x-11)\:>\:0$

It says: The product of two quantities is positive.

This is true if: [1] both factors are positive, or [2] both factors are negative.
. . And we must consider both cases.

Both positive: . $\begin{array}{ccc}x - 5 \:>\:0 & \Rightarrow & x \:> \:5 \\ x-11\:>\>0 & \Rightarrow & x \:>\:11\end{array}$

So $x$ must be greater than 5 and greater than 11.
. . Then: . $x \:>\:11$ covers both of them.

Both negative: . $\begin{array}{ccc}x - 5 \:<\:0 & \Rightarrow & x \:<\:5 \\ x-11 \:<\:0 & \Rightarrow & x \:<\:11 \end{array}$

So $x$ must be less than 5 and less than 11.
. . Then: . $x \:< \:5$ covers both of them.

Solution: . $x \:<\:5\:\text{ or }\:x\:>\:11$

Interval notation: . $(-\infty,\,5)\:\cup\11,\,\infty) " alt="(-\infty,\,5)\:\cup\11,\,\infty) " />

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Here's a rather primitive approach to the problem.

Consider the function: . $y \:=\x-5)(x-11)" alt="y \:=\x-5)(x-11)" />
. . When is this function positive?

We have a parabola $(y \:=\:x^2 - 16x + 55)$
. . When is it above the x-axis?

The parabola opens upward and it has x-intercepts 5 and 11.
The graph is below the x-axis between the intercepts
. . so it is above the x-axis to the left of 5 and to the right of 11.

See it?

3. I am not really understanding that. Well for example i have a somewhat similar problem of -x^2+4x>=0
for that i then did -x^2+4x-0 then i got (-x+2)(x+2) and my answe in interval was (-infinity,-2]U[-2,infinity) it says that was wrong, so would it be somewhat what you showed?

4. Originally Posted by Girlaaaaaaaa
1) How do i write the answer in interval notation for (x-5)(x-11)>0
$x^2 - 16x + 55 > 0 \implies (x - 8)^2 > 9 \implies \left| {x - 8} \right| > 3.$

Finally $- 3 > x - 8 > 3\,\therefore \,5 > x > 11.$

5. Originally Posted by Krizalid
Finally $- 3 > x - 8 > 3\,\therefore \,5 > x > 11.$
Excuse me? -3>x-8>3???