Solve the polynomial inequality and graph the solution set on a number line. Express the solution set in interval notation.

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- anonymous

x^2 - 4x - 12 ≤ 0

- ZeHanz

First, solve x²-4x-12=0, by factoring as (x ....)(x ....)=0.
Once you've found the zeroes, put these on the number line.
They will divide the number line into three pieces.
Because the graph of x²-4x-12 is a parabola with a minimum, the part of the number line between the two zeroes gives a negative value, the other parts a positive value.
Mark the areas with "+ + +", "- - -" and "+ + +".
Now you can see the solution of the inequality right in front of you: it is the middle part.

- anonymous

Im lost

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## More answers

- ZeHanz

Don't be! Can you factor x²-4x-12=0?

- anonymous

x*x - 2*2*x-2*2*3
Like that?

- ZeHanz

No, like (x + ...)(x + ...)=0. On the dots you need numbers a and b.
Now a+b=-4 (that's the coefficient of the middle term -4x),
and a*b = -12, the constant third term.
Can you find a and b?

- anonymous

2 & 12 being a & b?

- ZeHanz

No, because 2+12 = 14 instead of -4 and 2*12 = 24 instead of -12.
If you take a = 2 and b = -6, you get 2+(-6)=-4 and 2*(-6)=-12, so:
x²-4x-12=0 is the same as (x+2)(x-6)=0.
Are you familiar with this process?

- anonymous

Kinda,

- ZeHanz

OK, you probably need to practise this a few times, to get the hang of it.
Now look at (x+2)(x-6)=0. It is the product of two numbers, x+2 and x-6.
This product is 0.
If you multiply two numbers and the outcome is 0, what does that tell you about these numbers?

- anonymous

That the product is 0 still? I dont know..

- ZeHanz

If you have 4*6, 2*59, 63*456, you may not always be able to give the answer immediately, but one thing you do know: the answer is certainly NOT zero. Agreed?

- anonymous

Agreed,

- ZeHanz

So suppose I have multiplied two numbers and the outcome IS zero, what do you know about (at least one of) these numbers?

- anonymous

I dont get what your asking for, it would just be zero

- ZeHanz

It means that at least one of the numbers must be zero!
0*456=0, -34*0=0, 0*0=0, but 34*53 is not 0.
This means, if you have (x+2)(x-6)=0, you can split this equation into two very easy ones:
x+2=0 or x-6=0.
Can you see their solutions?

- anonymous

No

- ZeHanz

Can you solve the equation x+2=0?

- anonymous

-2

- ZeHanz

OK, and in the same way, the solution of x-6=0 is 6.
We now have the solutions of (x+2)(x-6)=0.
Solutions are -2 and 6.
Put them on a number line, like this:
------------|------------|-----------
-2 6

- anonymous

wouldnt it be -6? instead of just 6

- ZeHanz

No, because we need to set x=6 in the equation x-6=0 to make it true.
Next step:
We can also put zeroes above -2 and 6 to indicate the value of (x+2)(x-6) there:
0 0
------------|------------|-----------
-2 6

- ZeHanz

Keep in mind we are trying to solve x² - 4x - 12 ≤ 0.
We have indicated on the number line this: x² - 4x - 12 = 0.
That is not enough, we also need to know what values of x will make the thing < 0.

- ZeHanz

To find out what the sign (pos or neg) x²-4x-12 will get, just try a number between -2 and 6, e.g. x=0.
0² - 4*0 -12 = -12 <0, so negative there.
If you try in the other parts of the line you get this:

##### 1 Attachment

- anonymous

Wouldnt that make them infinitys?

- ZeHanz

If you put in very large numbers for x, the value of x^2 - 4x - 12 will also get very large, but we need not to worry about that, because we have to answer another question:
What values of x make x^2 - 4x - 12 smaller than zero?
The answer is visible in the image: between -2 and 6!
So if I color all solutions red, I get this:

##### 1 Attachment

- ZeHanz

Only one thing left: we have to write the solution (drawn in red) in interval notation...

- ZeHanz

How do you write all the numbers from -2 to 6 (including -2 and 6!) as an interval?

- anonymous

(-2,6)

- ZeHanz

That would exclude -2 and 6...

- ZeHanz

To include them, you write [-2, 6].
One final word of advice: from the answers you've given I get the impression that you are not yet ready to solve problems like this. I really think you should practise elementary algebra skills.
Nevertheless, I hope I have helped you a little...

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