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

Algebra ->  Quadratic Equations and Parabolas -> 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      Log 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) About Me  (Show Source):
You can put this solution on YOUR website!
# 1

x%5E2-3x=-10 Start with the given equation.


x%5E2-3x%2B10=0 Get all terms to the left side.


Notice we have a quadratic in the form of Ax%5E2%2BBx%2BC where A=1, B=-3, and C=10


Let's use the quadratic formula to solve for x


x+=+%28-B+%2B-+sqrt%28+B%5E2-4AC+%29%29%2F%282A%29 Start with the quadratic formula


x+=+%28-%28-3%29+%2B-+sqrt%28+%28-3%29%5E2-4%281%29%2810%29+%29%29%2F%282%281%29%29 Plug in A=1, B=-3, and C=10


x+=+%283+%2B-+sqrt%28+%28-3%29%5E2-4%281%29%2810%29+%29%29%2F%282%281%29%29 Negate -3 to get 3.


x+=+%283+%2B-+sqrt%28+9-4%281%29%2810%29+%29%29%2F%282%281%29%29 Square -3 to get 9.


x+=+%283+%2B-+sqrt%28+9-40+%29%29%2F%282%281%29%29 Multiply 4%281%29%2810%29 to get 40


x+=+%283+%2B-+sqrt%28+-31+%29%29%2F%282%281%29%29 Subtract 40 from 9 to get -31


x+=+%283+%2B-+sqrt%28+-31+%29%29%2F%282%29 Multiply 2 and 1 to get 2.


x+=+%283+%2B-+i%2Asqrt%2831%29%29%2F%282%29 Simplify the square root


x+=+%283%2Bi%2Asqrt%2831%29%29%2F%282%29 or x+=+%283-i%2Asqrt%2831%29%29%2F%282%29 Break up the expression.


So the answers are x+=+%283%2Bi%2Asqrt%2831%29%29%2F%282%29 or x+=+%283-i%2Asqrt%2831%29%29%2F%282%29



# 2

%28x%2B8%29%28x-8%29%28x%2B12%29%3E0 Start with the given inequality


%28x%2B8%29%28x-8%29%28x%2B12%29=0 Set the left side equal to zero


Set each individual factor equal to zero:

x%2B8=0, x-8=0 or x%2B12=0

Solve for x in each case:

x=-8, x=8 or x=-12


So our critical values are x=-8, x=8 and x=-12

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

number_line%28+600%2C+-15%2C+10%2C-8%2C8%2C-12%29



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 -12 (notice how it's to the left of the leftmost endpoint):

So let's pick x=-13

%28x%2B8%29%28x-8%29%28x%2B12%29%3E0 Start with the given inequality


%28-13%2B8%29%28-13-8%29%28-13%2B12%29%3E+0 Plug in x=-13


-105%3E+0 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.


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Let's pick a test value that is in between -12 and -8:

So let's pick x=-10

%28x%2B8%29%28x-8%29%28x%2B12%29%3E0 Start with the given inequality


%28-10%2B8%29%28-10-8%29%28-10%2B12%29%3E+0 Plug in x=-10


72%3E+0 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 ()





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Let's pick a test value that is in between -8 and 8:

So let's pick x=0

%28x%2B8%29%28x-8%29%28x%2B12%29%3E0 Start with the given inequality


%280%2B8%29%280-8%29%280%2B12%29%3E+0 Plug in x=0


-768%3E+0 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.


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Let's pick a test value that is greater than 8 (notice how it's to the right of the rightmost endpoint):

So let's pick x=9

%28x%2B8%29%28x-8%29%28x%2B12%29%3E0 Start with the given inequality


%289%2B8%29%289-8%29%289%2B12%29%3E+0 Plug in x=9


357%3E+0 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 ()





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Summary:

So the solution in interval notation is:


() ()





# 3

2x%5E2-5x%2B1 Start with the given expression.


2%28-4%29%5E2-5%28-4%29%2B1 Plug in x=-4.


2%2816%29-5%28-4%29%2B1 Square -4 to get 16.


32-5%28-4%29%2B1 Multiply 2 and 16 to get 32.


32%2B20%2B1 Multiply -5 and -4 to get 20.


53 Combine like terms.


So 2x%5E2-5x%2B1=53 when x=-4



# 4




%289x%5E8-3%29%28x%5E9-3%29 Start with the given expression.


Now let's FOIL the expression.


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


%28highlight%289x%5E8%29-3%29%28highlight%28x%5E9%29-3%29 Multiply the First terms:%289%2Ax%5E8%29%2A%28x%5E9%29=9%2Ax%5E17. Note: simply add the exponents


%28highlight%289x%5E8%29-3%29%28x%5E9%2Bhighlight%28-3%29%29 Multiply the Outer terms:%289%2Ax%5E8%29%2A%28-3%29=-27%2Ax%5E8.


%289x%5E8%2Bhighlight%28-3%29%29%28highlight%28x%5E9%29-3%29 Multiply the Inner terms:%28-3%29%2A%28x%5E9%29=-3%2Ax%5E9.


%289x%5E8%2Bhighlight%28-3%29%29%28x%5E9%2Bhighlight%28-3%29%29 Multiply the Last terms:%28-3%29%2A%28-3%29=9.


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So we have the terms: 9%2Ax%5E17, -27%2Ax%5E8, -3%2Ax%5E9, 9


9%2Ax%5E17-27%2Ax%5E8-3%2Ax%5E9%2B9 Now add every term listed above to make a single expression.


So %289x%5E8-3%29%28x%5E9-3%29 FOILs to 9%2Ax%5E17-27%2Ax%5E8-3%2Ax%5E9%2B9.


In other words, %289x%5E8-3%29%28x%5E9-3%29=9%2Ax%5E17-27%2Ax%5E8-3%2Ax%5E9%2B9.