- #1

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## Homework Statement

Question + attempt:

## Homework Equations

## The Attempt at a Solution

Why is it that when I expand the factored form, I don't get the original equation?

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- Thread starter ainster31
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- #1

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Question + attempt:

Why is it that when I expand the factored form, I don't get the original equation?

- #2

Mark44

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You get m## Homework Statement

Question + attempt:

## Homework Equations

## The Attempt at a Solution

Why is it that when I expand the factored form, I don't get the original equation?

- #3

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You get m^{2}+ (3/2)m + 1 = 0, which is equivalent to the original equation. Just multiply both sides by 2.

Let's say we have $$2{ m }^{ 2 }+3m+1=60$$.

Would I be able to apply the quadratic formula to the left hand-side of the equation only?

- #4

Mark44

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No. The Quadratic Formula requires that your equation be in the form axLet's say we have $$2{ m }^{ 2 }+3m+1=60$$.

Would I be able to apply the quadratic formula to the left hand-side of the equation only?

- #5

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How true is the below?

- #6

Mark44

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How true is the below?

What do you think? These look like textbook problems. Our policy here is that we don't do the work for students. Tell me what you think and why, and I'll tell you if you're right or where you went wrong.

- #7

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What do you think? These look like textbook problems. Our policy here is that we don't do the work for students. Tell me what you think and why, and I'll tell you if you're right or where you went wrong.

They're not problems. They're statements that I wrote.

If I recall correctly, in the 10th grade, the teacher said the factored form of a quadratic equation is ##y=(x-r)(x-s)## and in the 11th grade, the teacher said the factored form of a cubic equation is ##y=(x-r)(x-s)(x-t)##. However, in the 11th grade, the teacher changed it to ##y=a(x-r)(x-s)(x-t)## later.

I don't understand why the cubic equation has the a coefficient and not the quadratic one.

Last edited:

- #8

Mark44

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The general form of a quadraticThey're not problems. They're statements that I wrote.

If I recall correctly, in the 10th grade, the teacher said the general form of a quadratic equation is ##y=(x-r)(x-s)## and in the 11th grade, the teacher said the general form of a cubic equation is ##y=(x-r)(x-s)(x-t)##.

The general form of a cubic function is y = axHowever, in the 11th grade, the teacher changed it to ##y=a(x-r)(x-s)(x-t)## later.

- #9

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Edit: Actually, never mind. Thanks.

- #10

Mark44

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2m^{2} + 3m + 2 = 0

==> 2(m + 1)(m + 1/2) = 0

==> 2(m + 1)(m + 1/2) = 0

- #11

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Actually, my understanding still might be poor.

- #12

Mark44

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==> 2(x

==> 2(x - 1)(x + 1) = 0

The coefficient a that appears in ax

BTW, you can use the quadratic formula for this equation, but it's not necessary. After removing the common factor to get 2(x

For the problem in post #1, you have

2m

==> 2(m

==> 2(m + 1)(m + 1/2) = 0

To go from the 2nd step to the 3rd, I factored, but you can also use the quadratic formula.

Note that 2(m + 1)(m + 1/2) = 0 and (m + 1)(m + 1/2) = 0 are equivalent equations. They both have exactly the same solutions.

- #13

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Note that 2(m + 1)(m + 1/2) = 0 and (m + 1)(m + 1/2) = 0 are equivalent equations. They both have exactly the same solutions.

So what is the point of the a coefficient if you can always divide it out?

- #14

Mark44

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- #15

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My confession is that I'm a second-year computer engineering student.

It's kind of sad. Factoring quadratic and cubic equations always bugged me. The original quadratic is from trying to solve an auxiliary equation of a differential equation. Lol

- #16

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You need to multiply by 2, because when you were factoring, your formula was divided by 2.

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