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I can't solve a simple equation. Help please.

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SilverSolder:

--- Quote from: Terry Bites on February 05, 2021, 01:47:05 pm ---Just because its my birthday.... (Attachment Link)
MS mathematics- Handy, free, works... https://math.microsoft.com/en

--- End quote ---

This is actually rather good - a no BS application!  Remember those?

To download the last standalone desktop application:

https://www.microsoft.com/en-us/download/details.aspx?id=15702

You can say "No" to installing the ancient version of DirectX, with no ill effects

vad:
Another vote for Maxima / wxMaxima.

rstofer:
You can plug the equation in at www.symbolab.com, get both solutions.

IanB:

--- Quote from: intabits on February 02, 2021, 07:40:12 pm ---Thanks to all who replied.

Yes, I did manage to get to
ax^2 + axb - b = 0       

So I knew it was a quadratic, though I didn't think to divide by a to get
x^2 + bx - b/a = 0      

My usual method was to then try to factorize it into something like
(x - 1/a)(x + b) = 0       (incorrect)
and then find the roots by seeing when each of the two factors becomes zero.

I know that's not always possible, and that the quadratic formula was a method to handle that situation.
But although we were taught it, I can't ever recall actually using it!
So I didn't, hoping there was some simpler method.
Though that is (now) obviously the method to use.
--- End quote ---

Here is the way to do it if you can't remember the quadratic formula.

Start with the reduced quadratic:

x² + bx - b/a = 0

Now construct a similar equation that is easier to solve:

(x + b/2)² = 0

If we multiply this out, we get:

x² + bx + b²/4 = 0

We can make this similar equation the same as the equation we want like this:

x² + bx + b²/4 − b²/4 − b/a = x² + bx − b/a = 0

But now we are in a position to solve the original equation:

(x² + bx + b²/4) − b²/4 − b/a = (x + b/2)² − b²/4 − b/a = 0

We have:

(x + b/2)² − b²/4 − b/a = 0
(x + b/2)² = b²/4 + b/a
x + b/2 = ± √(b²/4 + b/a)

Therefore:

x = − b/2 ± √(b^2/4 + b/a)

This is the same answer as already presented above, and this method (called "completing the square") is how the quadratic formula is originally obtained.


intabits:
Thanks for the explanation. "completing the square" is a familiar term, but I'd long forgotten what it involved.

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