where did this term go?

allegansveritatem

Full Member
Joined
Jan 10, 2018
Messages
404
I was looking a method for deriving the quadratic formula from the form:ax^2+bx+c=0 but got taken up short when I couldn't account for the whereabouts of a term. Here is the book's presentation:
11510

Now, what is puzzling me is this: Where does the b/a times x go after the 4th equals sign? I mean., between the 4th and the fifth line of this proof, the b/a times x seems to fall out of the world. What am I missing. I tried several times to prove derive this formula for myself and came up with some really exotic expressions.
 

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
4,132
That's where they completed the square.

Look at it backward, starting from the LHS of the fifth line and expanding it. You'll get the LHS of the fourth line.

In effect, they are applying the fact that \(\displaystyle (a+b)^2 = a^2 + 2ab + b^2\), but in reverse: \(\displaystyle a^2 + 2ab + b^2 \rightarrow (a+b)^2\).
 

JeffM

Elite Member
Joined
Sep 14, 2012
Messages
3,565
I would do it a slightly different way.

Starting at line 3

\(\displaystyle x^2 + \dfrac{b}{a} * x = -\ \dfrac{c}{a} \implies\)

\(\displaystyle x^2 + 2 * \dfrac{b}{2a} * x = -\ \dfrac{c}{a} \implies\)

\(\displaystyle x^2 + 2 * \dfrac{b}{2a} * x + \left ( \dfrac{b}{2a} \right )^2 = \left ( \dfrac{b}{2a} \right )^2 - \dfrac{c}{a} \implies\)

\(\displaystyle \left(x + \dfrac{b}{2a} \right ) \left(x + \dfrac{b}{2a} \right ) = \left ( \dfrac{b}{2a} \right )^2 - \dfrac{c}{a} \implies \)

\(\displaystyle \left (x + \dfrac{b}{2a} \right )^2 = \left ( \dfrac{b}{2a} \right )^2 - \dfrac{c}{a} \implies\)

\(\displaystyle \left (x + \dfrac{b}{2a} \right )^2 = \dfrac{b^2}{4a^2} - \dfrac{c}{a} \implies\)

\(\displaystyle \left ( x + \dfrac{b}{2a} \right )^2 = \dfrac{b^2}{4a^2} - \dfrac{4ac}{4a^2} \implies\)

\(\displaystyle \left ( x + \dfrac{b}{2a} \right )^2 = \dfrac{b^2 - 4ac}{4a^2} \implies\)

\(\displaystyle x + \dfrac{b}{2a} = \pm \ \sqrt{\dfrac{b^2 - 4ac}{4a^2}} \implies\)

\(\displaystyle x + \dfrac{b}{2a} = \pm \ \dfrac{\sqrt{b^2 - 4ac}}{\sqrt{4a^2}} \implies\)

\(\displaystyle x + \dfrac{b}{2a} = \dfrac{\pm \ \sqrt{b^2 - 4ac}}{2a} \implies\)

\(\displaystyle x = \dfrac{-\ b \pm \sqrt{b^2 - 4ac}}{2a}.\)

It's the same logic involving \(\displaystyle (u + v)^2 = u^2 + 2uv + v^2\),

but we now see why we go \(\displaystyle \dfrac{b}{a} * x = \dfrac{2}{2} * \dfrac{b}{a} * x = 2 * \dfrac{b}{2a} * x.\)

That is to get the 2uv term. And we add \(\displaystyle \left ( \dfrac{b}{2a} \right )^2\)

to get the v^2 term.
 

allegansveritatem

Full Member
Joined
Jan 10, 2018
Messages
404
That's where they completed the square.

Look at it backward, starting from the LHS of the fifth line and expanding it. You'll get the LHS of the fourth line.

In effect, they are applying the fact that \(\displaystyle (a+b)^2 = a^2 + 2ab + b^2\), but in reverse: \(\displaystyle a^2 + 2ab + b^2 \rightarrow (a+b)^2\).
I don't have a pen with me or I would immediately check what you say here. But, eyeballing it, I think you are on to it. Thanks
 

allegansveritatem

Full Member
Joined
Jan 10, 2018
Messages
404
I would do it a slightly different way.

Starting at line 3

\(\displaystyle x^2 + \dfrac{b}{a} * x = -\ \dfrac{c}{a} \implies\)

\(\displaystyle x^2 + 2 * \dfrac{b}{2a} * x = -\ \dfrac{c}{a} \implies\)

\(\displaystyle x^2 + 2 * \dfrac{b}{2a} * x + \left ( \dfrac{b}{2a} \right )^2 = \left ( \dfrac{b}{2a} \right )^2 - \dfrac{c}{a} \implies\)

\(\displaystyle \left(x + \dfrac{b}{2a} \right ) \left(x + \dfrac{b}{2a} \right ) = \left ( \dfrac{b}{2a} \right )^2 - \dfrac{c}{a} \implies \)

\(\displaystyle \left (x + \dfrac{b}{2a} \right )^2 = \left ( \dfrac{b}{2a} \right )^2 - \dfrac{c}{a} \implies\)

\(\displaystyle \left (x + \dfrac{b}{2a} \right )^2 = \dfrac{b^2}{4a^2} - \dfrac{c}{a} \implies\)

\(\displaystyle \left ( x + \dfrac{b}{2a} \right )^2 = \dfrac{b^2}{4a^2} - \dfrac{4ac}{4a^2} \implies\)

\(\displaystyle \left ( x + \dfrac{b}{2a} \right )^2 = \dfrac{b^2 - 4ac}{4a^2} \implies\)

\(\displaystyle x + \dfrac{b}{2a} = \pm \ \sqrt{\dfrac{b^2 - 4ac}{4a^2}} \implies\)

\(\displaystyle x + \dfrac{b}{2a} = \pm \ \dfrac{\sqrt{b^2 - 4ac}}{\sqrt{4a^2}} \implies\)

\(\displaystyle x + \dfrac{b}{2a} = \dfrac{\pm \ \sqrt{b^2 - 4ac}}{2a} \implies\)

\(\displaystyle x = \dfrac{-\ b \pm \sqrt{b^2 - 4ac}}{2a}.\)

It's the same logic involving \(\displaystyle (u + v)^2 = u^2 + 2uv + v^2\),

but we now see why we go \(\displaystyle \dfrac{b}{a} * x = \dfrac{2}{2} * \dfrac{b}{a} * x = 2 * \dfrac{b}{2a} * x.\)

That is to get the 2uv term. And we add \(\displaystyle \left ( \dfrac{b}{2a} \right )^2\)

to get the v^2 term.
very pretty. I think my problem, as Dr. P points out above, has to do with the fact that the LHS is a perfect square and therefore expressible in the way the book expressed it.
 

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
4,132
Have you learned how to complete the square in a case with numerical coefficients, like x^2 - 6x + 4 = 0? That's an important skill, and is what they used here with parameters. I don't know that the asterisk in your original image refers to, but you might want to check it out; if you skipped over that section of the book earlier, be sure to go back there and master it. You will be seeing it again, and again!

Even more generally, you need to be able to recognize a perfect square when you see it.
 

allegansveritatem

Full Member
Joined
Jan 10, 2018
Messages
404
Have you learned how to complete the square in a case with numerical coefficients, like x^2 - 6x + 4 = 0? That's an important skill, and is what they used here with parameters. I don't know that the asterisk in your original image refers to, but you might want to check it out; if you skipped over that section of the book earlier, be sure to go back there and master it. You will be seeing it again, and again!

Even more generally, you need to be able to recognize a perfect square when you see it.
I NEVER skip anything. I do know about completing the square and have used the method many times...but sometimes I go technique-blind.
 

Denis

Senior Member
Joined
Feb 17, 2004
Messages
1,724
Good...I'm proud of you :)
 
Top