SOLUTION: I have a few problems I'm struggling with: 1: x^2 - 3x = -10 (directions say use quadratic formula, use radicals as needed). 2: (x+12)(x-8)(x+8)>0 (directions say to use on

Question 196896: I have a few problems I'm struggling with:
1: x^2 - 3x = -10 (directions say use quadratic formula, use radicals as needed).
2: (x+12)(x-8)(x+8)>0 (directions say to use one equality or compount equality)
3: 2x^2 - 5x +1 (evaluate polynomial for x= -4
4: (9x^8 - 3)(x^9 - 3) (use FOIL to find product)
I thank you in advance for time spent in helping with these problems
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
# 1

Start with the given equation.

Get all terms to the left side.

Notice we have a quadratic in the form of where , , and

Let's use the quadratic formula to solve for x

Start with the quadratic formula

Plug in , , and

Negate to get .

Square to get .

Multiply to get

Subtract from to get

Multiply and to get .

Simplify the square root

or Break up the expression.

So the answers are or

# 2

Start with the given inequality

Set the left side equal to zero

Set each individual factor equal to zero:

, or

Solve for x in each case:

, or

So our critical values are , and

Now set up a number line and plot the critical values on the number line

So let's pick some test points that are near the critical values and evaluate them.

Let's pick a test value that is less than (notice how it's to the left of the leftmost endpoint):

So let's pick

Start with the given inequality

Plug in

Evaluate and simplify the left side

Since the inequality is false, this means that the interval does not work. So this interval is not in our solution set and we can ignore it.

---------------------------------------------------------------------------------------------

Let's pick a test value that is in between and :

So let's pick

Start with the given inequality

Plug in

Evaluate and simplify the left side

Since the inequality is true, this means that the interval works. So this tells us that this interval is in our solution set.
So part our solution in interval notation is (

)

---------------------------------------------------------------------------------------------

Let's pick a test value that is in between and :

So let's pick

Start with the given inequality

Plug in

Evaluate and simplify the left side

Since the inequality is false, this means that the interval does not work. So this interval is not in our solution set and we can ignore it.

---------------------------------------------------------------------------------------------

Let's pick a test value that is greater than (notice how it's to the right of the rightmost endpoint):

So let's pick

Start with the given inequality

Plug in

Evaluate and simplify the left side

Since the inequality is true, this means that the interval works. So this tells us that this interval is in our solution set.
So part our solution in interval notation is (

)

---------------------------------------------------------------------------------------------

Summary:

So the solution in interval notation is:

(

)

(

)

# 3

Start with the given expression.

Plug in .

Square to get .

Multiply and to get .

Multiply and to get .

Combine like terms.

So when

# 4

Start with the given expression.

Now let's FOIL the expression.

Remember, when you FOIL an expression, you follow this procedure:

Multiply the First terms:. Note: simply add the exponents

Multiply the Outer terms:.

Multiply the Inner terms:.

Multiply the Last terms:.

---------------------------------------------------
So we have the terms: , , ,

Now add every term listed above to make a single expression.

So FOILs to .

In other words, .

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