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Quadratic Formula: Solve for b


Date: 08/15/2001 at 11:29:02
From: Shawn
Subject: Quadratic Formula

How do you solve for b in the quadratic formula?


Date: 08/15/2001 at 12:48:28
From: Doctor Ian
Subject: Re: Quadratic Formula

Hi Shawn,

I'm not entirely sure why you'd _want_ to solve for b, since it's 
normally a constant, but here's how you would do it.

The quadratic formula,

                   -b +/- sqrt(b^2 - 4ac)
              x = -----------------------
                         2a

applies whenever you have an equation like 

   ax^2 + bx + c = 0

So we can start with either the formula, or the equation. Frankly, the
equation looks as if it will be less work:

   ax^2 + bx + c = 0

              bx = -ax^2 - c  

               b = (-ax^2 - c) / x

Is this what you wanted to know? 

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/   


Date: 08/15/2001 at 13:33:12
From: Shawn Lilianstrom
Subject: Re: Quadratic Formula

No, you didn't answer my question. One of my summer math problems is 
"to solve for b in the quadratic formula." I don't know how to do it.


Date: 08/16/2001 at 15:01:05
From: Doctor Ian
Subject: Re: Quadratic Formula

Hi Shawn,

I _did_ solve for b.  The standard quadratic form

  ax^2 + bx + c = 0

is exactly equivalent to the quadratic formula

      -b +/- sqrt(b^2 - 4ac)
  x = ----------------------
             2a

So you could 'undo' the formula, and then proceed as I proceeded. 

  
      -b +/- sqrt(b^2 - 4ac)
  x = ----------------------
             2a

   .
   .  
   .

  ax^2 + bx + c = 0

             bx = -ax^2 - c

              b = (-ax^2 - c) / x

But since you _know_ that there is a path from the formula to the 
equation, there is no need for you to write it down. 

Now, if you mean that you need to work from the actual formula to an
equation like

  b = ...?

then here is how you would start:

                 -b +/- sqrt(b^2 - 4ac)
             x = ----------------------
                        2a

           2ax = -b +/- sqrt(b^2 - 4ac)

       2ax + b = +/- sqrt(b^2 - 4ac)

   (2ax + b)^2 = b^2 - 4ac

That's the tricky part. Now you just expand both sides, cancel out the
extra 4's and a's, and do the usual algebraic trickery. 

It's important to realize that the first solution that I gave you is 
the "mathematician's" solution. The following joke illustrates the 
point I was trying to make:

  A mathematician and a physicist were arguing over whose field of 
  study was better. They decided to settle the argument by posing 
  questions. The mathematician went first, and posed a complicated
  mathematical problem. With a great deal of effort, several books 
  of mathematical tables and techniques, and a few hours, the  
  physicist gave the solved problem to the mathematician, who was duly 
  impressed. 

  "All right, my turn. Here's the problem: you have a pot of water 
  on the stove, at 60 F. You want to heat it up to 70 F. What do you
  do?" The mathematician replied, "Oh, that's easy. You turn the stove 
  on. Fourier's equations govern how heat transfers from the stove to 
  the pot, and you can solve them numerically to find out how long it 
  takes for the water to reach 70 F." The physicist then asks, "All 
  right, so what if the water's at 65 F?" 

  "Oh, that's even easier. You take the pot of water, stick it in the
  refrigerator until it cools down to 60 F, and then it simplifies to 
  the previous problem!" 

Here is a slightly different version of the same joke:

  A physicist and a mathematician are in the faculty lounge having 
  a cup of coffee when, for no apparent reason, the coffee machine 
  bursts into flames. The physicist rushes over to the wall, grabs
  a fire extinguisher, and fights the fire successfully. The same 
  time next week, the same pair are there drinking coffee and talking 
  shop when the new coffee machine catches on fire. The mathematician 
  stands up, fetches the fire extinguisher, and hands it to the 
  physicist, thereby reducing the problem to one already solved... 

In another variation on the second joke, the mathematician realizes 
that the previous week's episode shows that a solution to the problem 
exists, so he ignores it. 

Now, these are funny, but they are funny _because_ they expose a deep 
truth about the way mathematicians do mathematics.

By noticing that the standard quadratic equation and the quadratic 
formula are equivalent, I saw that I could reduce either one to the 
other, which meant that I could choose either as my starting point, 
knowing that a conversion between the two existed. Knowing that it 
existed, there was no point in my reproducing it. 

I bring this up only because one of the ways to make your math classes 
more entertaining is to try to approach them the way a mathematician 
would. Now, it's easy to mistakenly think that this means you should 
learn to enjoy jotting down pages and pages of equations. But what I'm 
trying to get you to see is that what it _really_ means that you 
should get into the habit of looking for the easiest possible way to 
get from the statement of a problem to its solution... which in many 
cases means finding a short connection between the problem you're 
working on and one that you know has already been solved, whether or 
not you were the one who solved it. 

I hope this helps.  Write back if you'd like to talk about this some 
more, or if you have any other questions. 

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School About Math
High School Basic Algebra

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