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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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