1. Oct 15, 2013

### ainster31

1. The problem statement, all variables and given/known data

Question + attempt:

2. Relevant equations

3. The attempt at a solution

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

2. Oct 15, 2013

### Staff: Mentor

You get m2 + (3/2)m + 1 = 0, which is equivalent to the original equation. Just multiply both sides by 2.

3. Oct 15, 2013

### ainster31

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. Oct 15, 2013

### Staff: Mentor

No. The Quadratic Formula requires that your equation be in the form ax2 + bx + c = 0, with a ≠ 0.

5. Oct 15, 2013

### ainster31

How true is the below?

6. Oct 15, 2013

### Staff: Mentor

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. Oct 15, 2013

### ainster31

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: Oct 15, 2013
8. Oct 15, 2013

### Staff: Mentor

The general form of a quadratic function is y = ax2 + bx + c. The right side of this can be written as a(x -r)(x -s). It's possible that r and s are complex numbers, though.
The general form of a cubic function is y = ax3 + bx2 + cx + d. The right side can be written in factored form as a(x - r)(x - s)(x - t), with possibly some of the roots being complex. Complex roots come in pairs, so we can't have more than two complex roots in a cubic equation.

9. Oct 15, 2013

### ainster31

Is there an example of a general quadratic equation where when factored, a is not 1?

Edit: Actually, never mind. Thanks.

10. Oct 15, 2013

### Staff: Mentor

2m2 + 3m + 2 = 0
==> 2(m + 1)(m + 1/2) = 0

11. Oct 15, 2013

### ainster31

Actually, my understanding still might be poor.

12. Oct 15, 2013

### Staff: Mentor

2x2 - 2 = 0
==> 2(x2 - 1) = 0
==> 2(x - 1)(x + 1) = 0

The coefficient a that appears in ax2 + bx + c = 0 is exactly the same as the one that appears in a(x - r)(x - s).

BTW, you can use the quadratic formula for this equation, but it's not necessary. After removing the common factor to get 2(x2 - 1), I just factored the expression inside the parentheses, using the formula x2 - b2 = (x - b)(x + b).

For the problem in post #1, you have
2m2 + 3m + 2 = 0
==> 2(m2 + (3/2)m + 1) = 0
==> 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. Oct 15, 2013

### ainster31

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

14. Oct 15, 2013

### Staff: Mentor

There's not really a point. a is just the coefficient of the squared term in a quadratic equation, just like be is the coefficient of the first degree term, and c is the constant.

15. Oct 15, 2013

### ainster31

Thanks for the help.

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. Oct 15, 2013

### Martin Zhao

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