From Förberedande kurs i matematik 1

Revision as of 06:47, 21 August 2008 (edit) Tekbot (Talk | contribs) m (Robot: Automated text replacement (-[[Bild: +[[Image:)) ← Previous diff		Current revision (10:18, 29 September 2008) (edit) (undo) Tek (Talk | contribs) m
(2 intermediate revisions not shown.)
Line 1:		Line 1:
-	{{NAVCONTENT_START}}	+	A first thought is perhaps to write the equation as
-	<center> [[Image:2_3_4a-1(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>x^{2}+ax+b=0</math>}}
-	{{NAVCONTENT_START}}	+
-	<center> [[Image:2_3_4a-2(2).gif]] </center>	+	and then try to choose the constants ''a'' and ''b'' in some way so that
-	{{NAVCONTENT_STOP}}	+	<math>x=-1</math> and <math>x=2</math> are solutions. But a better way is to start with a factorized form of a second-order equation,
		+
		+	{{Displayed math||<math>(x+1)(x-2)=0\,\textrm{.}</math>}}
		+
		+	If we consider this equation, we see that both <math>x=-1</math> and <math>x=2</math> are solutions to the equation, since <math>x=-1</math> makes the first factor on the left-hand side zero, whilst <math>x=2</math> makes the second factor zero. Also, it really is a second order equation, because if we multiply out the left-hand side, we get
		+
		+	{{Displayed math||<math>x^{2}-x-2=0\,\textrm{.}</math>}}
		+
		+	One answer is thus the equation <math>(x+1)(x-2)=0</math>, or <math>x^{2}-x-2=0\,</math>.
		+
		+
		+	Note: There are actually many answers to this exercise, but what all second-degree equations that have <math>x=-1</math> and <math>x=2</math> as roots have in common is that they can be written in the form
		+
		+	{{Displayed math||<math>ax^{2}-ax-2a=0\,,</math>}}
		+
		+	where ''a'' is a non-zero constant.

---

Current revision

A first thought is perhaps to write the equation as

\displaystyle x^{2}+ax+b=0

and then try to choose the constants a and b in some way so that \displaystyle x=-1 and \displaystyle x=2 are solutions. But a better way is to start with a factorized form of a second-order equation,

\displaystyle (x+1)(x-2)=0\,\textrm{.}

If we consider this equation, we see that both \displaystyle x=-1 and \displaystyle x=2 are solutions to the equation, since \displaystyle x=-1 makes the first factor on the left-hand side zero, whilst \displaystyle x=2 makes the second factor zero. Also, it really is a second order equation, because if we multiply out the left-hand side, we get

\displaystyle x^{2}-x-2=0\,\textrm{.}

One answer is thus the equation \displaystyle (x+1)(x-2)=0, or \displaystyle x^{2}-x-2=0\,.

Note: There are actually many answers to this exercise, but what all second-degree equations that have \displaystyle x=-1 and \displaystyle x=2 as roots have in common is that they can be written in the form

\displaystyle ax^{2}-ax-2a=0\,,

where a is a non-zero constant.