Talk:Algebra/Solving equations

Weda's theorem? i hear Vieta's theorem. http://mathworld.wolfram.com/VietasFormulas.html

Changing chapter name to "Solving quadratic equations"
I'm thinking of changing this chapter's name to Algebra:Solving quadratic equations because basically that's what it's about. Any feedback on this idea? H Padleckas 19:49, 10 November 2005 (UTC)

Perhaps some other algebra topics will be added later, but certainly Solving Quadratic Equations should be a topic. The first method described should be the graphing one since students all have graphing calculators and are tuned in to them. I mean you solve ax2 + bx + c = 0 by graphing y = ax2 + bx + c and looking for its x intercepts.

For the factoring method, I would like to do the same example or a more complicated one by the method of listing factors of the first and 3rd terms under the terms and cross multiplying to check. Many students find this method of organizing the guesses and tests very helpful! I did the example in a .jpg file but do not have permission to upload it here (I'm a Wikiversity contributor in good standing but not a wikibooks contributor yet - don't really understand why we have some topics in one and others in the other). Harveybrown (talk) 02:31, 10 May 2008 (UTC)

Solutions
I think it's very weird that the solutions are directly visible after the example. Maybe its a better idea to but them at the bottom of the page?
 * Solutions following examples are OK, but there should be "homework" type exercises written with the solutions (perhaps) on a separate page. H Padleckas 17:53, 24 November 2005 (UTC)

Merge with Algebra:Quadratic functions
There seems to be a lot of overlap with Algebra:Quadratic functions.

Weda
I too send this appeal to change the name Weda's theorem to Viete's formulas. Every high school student knows this.

correction
I have a correction. I will ask you to do it, because I am not signed, and such edit will be considered destruction. OK: it's written: the equation has two roots with different signs if $$\Delta\ge0,\ \frac{c}{a}<0\ \mbox{or}\ ac<0$$ now, if ac is negative, then (-4ac) will be positive, and the descriminant will be surely positive. therefore, if one needs to find for what values the equation has two roots with different signs, he needs to check when is ac<0, and only this (because for this values the discriminant will be positive for sure). I hope you understood me. If you didn't, write and i'll try to explain myself more clearly. Yours, from Israel. I am avilable here: he:משתמש:Hbk3 if someone saw this, please comment.

Completing the square
Quote:

$$   \begin{matrix} x^2+8x+9&=&0\\ x^2+8x&=&-9\\ x^2+8x+4^2&=&4^2-9\\ (x+4)^2&=&7\\ &x+4=\sqrt{7}& \quad \mbox{or} \quad x+4=-\sqrt{7}\\ &x=-4+\sqrt{7}& \quad \mbox{or} \quad x=-4-\sqrt{7}\\ &x=-1.35& \quad \mbox{or} \quad x=-6.65 \end{matrix} $$

In general, we get

$$x^2+kx+\left({k \over 2}\right)^2=\left(x+{k \over 2}\right)^2 $$

How does this general formula relate to the example ???? Where has the -9 gone to ?? It is at least quite confusing.

Yes I know that (a + b) x (a + b) = a2 + 2ab + b2. I learned that 65 years ago.

Andre anckaert (talk) 13:35, 11 January 2010 (UTC)


 * As a way to advance the solution $$4^2$$ (your proposed $$b^2$$) was added (and the complement on the other side so it balances as 0, but since it is square the signal is ignored) that is how -9 becomes 7. --Panic (talk) 20:34, 15 January 2010 (UTC)