## Question on deriving the quadratic formula

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
geramul
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### Question on deriving the quadratic formula

I have only one question for this.

Let me write out where I got stuck.

(x + b/2a)2 = -c/a + b2/4a2

What I can't figure out is how the common denominator of 4a is gotten. Let me give an example. Lets say we want to add together 1/4 and 5/8. We would look of course at the multiples of 4 and find one that's equal to other denominator. So we find that 8 is the least common denominator for both fractions. We multiply 1/4 by 2/2 to get 2/8, and now we can solve.

Now lets look at -c/a. Any variable without a coefficient has an implied coefficient of 1 yes? So really we can look at -c/a as -1c/1a. How can we get a common denominator of 4 from 1, when 1 only has itself as multiple? This doesn't make sense to me. The only way I could figure it out was if both bases were the same, we could just say the fraction was equal to 1 and turn that into 1/2, therefore giving us the ability to get 4. Someone please explain this to me.

stapel_eliz
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Where did you get stuck? What was the line before the one you posted? What was the last line that you understood?

Thank you.

geramul
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### Re: Question on deriving the quadratic formula

I said where I got stuck in my original post.

If you want me to write out the entire thing though.

1. ax2 + bx + c = 0

2. ax2 + bx = -c

3. x2 + bx/a = -c/a

4. x2 + bx/a + b2/4a2 = -c/a + b2/4a2

5. (x + b/2a)2 = -c/a + b2/4a2

This is where I'm stuck. How do you get 4a/4a from the a in -c/a? Like I said in my original post, a variable without a coefficient has an implied coefficient of 1. So really -c/a is -1c/1a. How does 4a come from 1a? 1 only has itself as a multiple, where is the 4 coming from?

stapel_eliz
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### Re: Question on deriving the quadratic formula

geramul wrote:4. x2 + bx/a + b2/4a2 = -c/a + b2/4a2

5. (x + b/2a)2 = -c/a + b2/4a2

This is where I'm stuck. How do you get 4a/4a from the a in -c/a?

I'm not seeing a "4a/4a", and the "-c/a" appears to be unchanged between line 4 and line 5...?

Perhaps it would help if you reviewed this page which shows the step-by-step derivation of the Quadratic Formula, using the completing-the-square methodology, and then replied saying where in that process the explanation stops making sense. Thank you!

strumbore
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### Re: Question on deriving the quadratic formula

everything got divided through by "a" (step 4), which leaves the whole equation essentially unchanged.

You just didn't notice, that's all.

After step 5, which is not shown in your post, multiply the term "-c/a" by "4a/4a", which will give you "-4ac/4a2"

Add the two terms on the right, square-root both sides (+/- when you do that), and bring "b/2a" over to the right side too.

Add those together and you're done.

x = -b +- SQRT(b2 - 4ac) all-over-2a

jg.allinsymbols
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### Re: Question on deriving the quadratic formula

$\frac{{ - c}}{a} \cdot \frac{{4a}}{{4a}} = \frac{{ - 4ac}}{{4a^2 }}$