Determining Values for A, B, and C in Quadratic Formula

TiaharaJBennett

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I have a question:

For the quadratic equation (2x+1)(x+3)=0, does 2=a, 1=b, and 3=c?
 
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srmichael

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Determining Values for A,B, and C in Quadratic Formula

I have a question:

For the quadratic equation (2x+1)(x+3)=0, does 2=a, 1=b, and 3=c?
FIRST, please start a new thread for each new probelem you have.

No. The standard form of a quadratic equation is y = ax^2 + bx + c so you will first need to multiply (2x+1)(x+3) out and get it into standard form. Then you can see what a, b and c are.
 
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TiaharaJBennett

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FIRST, please start a new thread for each new probelem you have.

you will first need to multiple (2x+1)(x+3) out and get it into standard form. Then you can see what a, b and c are.
You mean multiply?
 

mmm4444bot

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Of course he means multiply! Have you not learned FOIL yet?! :cool:

PS: I split your previous thread. Note the useful subject line.
 

TiaharaJBennett

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Of course he means multiply! Have you not learned FOIL yet?! :cool:

PS: I split your previous thread. Note the useful subject line.
Yes, I know about FOIL. But I haven't used it since I was in 9th grade. I'll figure it out though.
 
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mmm4444bot

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What grade are you in now?
 

TiaharaJBennett

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Okay, I did the FOIL method, and got:
2x2+7x+3 :)

2=a,7=b, and 3=c

Right? Please say yes.
 

mmm4444bot

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Yes.

Are you thinking that you need to use the Quadratic Formula to solve the equation (2x+1)(x+3) = 0 ?

If so, I would like you to know that there is a much simpler method that uses something called the Zero-Product Property.
 

TiaharaJBennett

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Yes.

Are you thinking that you need to use the Quadratic Formula to solve the equation (2x+1)(x+3) = 0 ?

If so, I would like you to know that there is a much simpler method that uses something called the Zero-Product Property.
I've never learned that. How does it work? But yeah, I was going to just use the quadratic formula.
 
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mmm4444bot

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I think that you should continue using the Quadratic Formula and then post your results.

Then, I will explain how to apply the Zero-Product Property, and you can try it again as practice/check. :cool:
 

TiaharaJBennett

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I think that you should continue using the Quadratic Formula and then post your results.

Then, I will explain how to apply the Zero-Product Property, and you can try it again as practice/check. :cool:
Okay, will do! :)
 

TiaharaJBennett

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I'll just post my work. I want to make sure I did this right.

x=-(7)±√ (7)2-4(2)(3)
___________________
2(2)

-(7)±√ 49-24 -7±√25
____________ = _______
4 4


=-7±5 -7-5
________ = _______
4 4


12/4, -2/2

x=-3, x=-1
 

mmm4444bot

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12/4, -2/2




Your work seems correct through the following result.

\(\displaystyle \frac{-7 \pm 5}{4}\)

After that, double-check your arithmetic when adding 5 to -7 and when subtracting 5 from -7.
 

TiaharaJBennett

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I did. I'm still getting the same thing.
 

mmm4444bot

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Okay, then check the typing in your post. Did you make typographical errors? You wrote:

12/4, -2/2
 

TiaharaJBennett

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mmm4444bot

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:idea: There is a preview button next to the submit button for proofreading posts before submission.




How did (-7 + 5)/4 turn into -2/2 ?
 

TiaharaJBennett

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How did (-7 + 5)/4 turn into -2/2 ?
-7+5=-2 and I got -2/2.

I see what I did wrong. That was supposed to be -7+5/4 to get -0.5, right?
 

mmm4444bot

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-7+5=-2 and I got -2/2.

That was supposed to be (-7+5)/4 to get -0.5, right?
Correct.

Btw, you may report x = -1/2, even though your calculator displays -0.5 (CASE-IN-POINT: Some students who consistently avoid working with small fractions eventually lose their ability to work with small fractions.)

:idea: You can always check your candidates for solutions by substituting them (one at a time) into the original equation, followed by doing all of the resulting arithmetic, to ensure that they work by demonstrating that you end up with a true statement (like 0 = 0 or 22/7 = 22/7 or -5=-5).



The Zero-Product Property tells us that anytime you get zero after multiplying things together, at least one of those things must be zero. In other words, it is impossible to get zero as a product when multiplying non-zero numbers together.

EGs:

Given a*b = 0

We can tell just by looking at this equation that either
a=0 or b=0 or both a and b equal zero.

We can't be sure which of these three scenarios is true without further information, but the Zero-Product Property tells us that ONE OF THOSE POSSIBILITIES must be true. So, we answer
a = 0 OR b = 0 OR both equal zero to cover all possibilities.


Given (x - 4)(3x + 7) = 0

We can tell just by looking that IF the factor
x - 4 = 0 then the product on the left side will equal zero because (0)(3x + 7) = 0.

Same result IF factor
3x + 7 = 0 because (x - 4)(0) = 0.


The examples above show why zero is such a powerful number. Multiplication by zero always results in zero.

What are the solutions to the equation (x - 4)(3x + 7) = 0 ?

Well, on the left side of the equation, the two unknown numbers (expressed above as x-4 and 3x+7) must multiply together to make zero. We already know that if just one of the two numbers is zero, then it does not matter what the other number is. Right?

Let's assume, one by one, that either factor is zero. Either of them could be; after all, their values are unknown to us at this point.

So, we apply the Zero-Product Property and write:

x - 4 = 0

OR

3x + 7 = 0

The solutions to these two equations (obtained by assuming that each factor is zero) will be the same as the solutions to the original equation.

This is because a value for the variable x=4 makes the expression x-4 zero AND a value x=-7/3 causes the number 3x+7 to evaluate to zero.

Check it out.

(x - 4)(3x + 7) = 0

(4 - 4)(12 + 7) = 0

(0)(19) = 0

0 = 0

(x - 4)(x*3 + 7) = 0

(-7/3 - 4)(-7/3 * 3 + 7) = 0

(-7/3 - 12/3)(-7 + 7) = 0

(-19/3)(0) = 0

0 = 0

Both assumed solutions check, and we know that a quadratic equation may not have more than two different solutions. Hence, the solutions to the original equation are:

x = 4 OR x = -7/3

Now, try using the Zero-Product Property on your exercise, and check that you get the same results as you did using the Quadratic Formula. Which method is easier for you? :cool: (DISCLAIMER: When an instructor specifically instructs to use the Quadratic Formula for a particular exercise, then one must demonstrate the Quadratic Formula.)
 
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