2.3 Quadratic expressions - Revision history

Phil: Edited for UK idiom, notation and terminology

2008-12-31T19:03:31Z

Edited for UK idiom, notation and terminology

←Older revision		Revision as of 19:03, 31 December 2008
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	{{Info\|		{{Info\|
	'''Contents:'''		'''Contents:'''
-	*Completing the square ~~method~~	+	*Completing the square
	*Quadratic equations		*Quadratic equations
	* Factorising		* Factorising
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	After this section, you will have learned to:		After this section, you will have learned to:
-	*Complete the square ~~for~~ expressions ~~of degree two (second degree)~~.	+	*Complete the square in quadratic expressions.
	*Solve quadratic equations by completing the square (not using a standard formula) and know how to check the answer.		*Solve quadratic equations by completing the square (not using a standard formula) and know how to check the answer.
-	*Factorise ~~second degree~~ expressions (when possible).	+	*Factorise quadratic expressions (when possible).
	*Directly solve factorised or almost factorised quadratic equations.		*Directly solve factorised or almost factorised quadratic equations.
-	*Determine the minimum / maximum value of an expression ~~of degree two~~.	+	*Determine the minimum / maximum value of a quadratic expression.
-	*Sketch parabolas by completing the square ~~method~~.	+	*Sketch parabolas by completing the square.
	}}		}}

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-	~~Simpler~~ forms of quadratic equations can be solved directly by taking roots.	+	The simplest forms of quadratic equations can be solved directly by taking roots.
	<div class="regel">		<div class="regel">
	The equation <math>x^2=a</math> where <math>a</math> is a positive number has two solutions (roots) <math>x=\sqrt{a}</math> and <math>x=-\sqrt{a}</math>.		The equation <math>x^2=a</math> where <math>a</math> is a positive number has two solutions (roots) <math>x=\sqrt{a}</math> and <math>x=-\sqrt{a}</math>.
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	*<math>x-1 = -\sqrt{16} = -4\,</math> which gives <math>x=1-4=-3</math>. </li>		*<math>x-1 = -\sqrt{16} = -4\,</math> which gives <math>x=1-4=-3</math>. </li>
	<li> Solve the equation <math>\ 2(x+1)^2 -8=0</math>. <br><br>		<li> Solve the equation <math>\ 2(x+1)^2 -8=0</math>. <br><br>
-	Move the term <math>8</math> over to the ~~righthand~~ side and divide both sides by <math>2</math>,	+	Move the term <math>8</math> over to the right-hand side and divide both sides by <math>2</math>,
	{{Displayed math\|\|<math>(x+1)^2=4 \; \mbox{.}</math>}}		{{Displayed math\|\|<math>(x+1)^2=4 \; \mbox{.}</math>}}
	Taking the roots gives:		Taking the roots gives:
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	</div>		</div>

-	To solve a quadratic equation generally we use a technique called completing the square.	+	To solve a quadratic equation more generally we can use a technique called completing the square.

	If we consider the rule for expanding a quadratic,		If we consider the rule for expanding a quadratic,
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	Divide both sides by 2		Divide both sides by 2
	{{Displayed math\|\|<math>x^2-x-\textstyle\frac{3}{4}=0\mbox{.}</math>}}		{{Displayed math\|\|<math>x^2-x-\textstyle\frac{3}{4}=0\mbox{.}</math>}}
-	Complete the square of the ~~lefthand~~ side (use <math>a=-\tfrac{1}{2}</math>)	+	Complete the square on the left-hand side (use <math>a=-\tfrac{1}{2}</math>)
	{{Displayed math\|\|<math>\textstyle\underline{\vphantom{\bigl(\frac{3}{4}}x^2-x} -\frac{3}{4} = \underline{\bigl(x-\frac{1}{2}\bigr)^2 - \bigl(-\frac{1}{2}\bigr)^2} -\frac{3}{4}= \bigl(x-\frac{1}{2}\bigr)^2 -1</math>}}		{{Displayed math\|\|<math>\textstyle\underline{\vphantom{\bigl(\frac{3}{4}}x^2-x} -\frac{3}{4} = \underline{\bigl(x-\frac{1}{2}\bigr)^2 - \bigl(-\frac{1}{2}\bigr)^2} -\frac{3}{4}= \bigl(x-\frac{1}{2}\bigr)^2 -1</math>}}
	This gives us the equation		This gives us the equation
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	'''Hint: '''		'''Hint: '''

-	Keep in mind that we can always test our solution to an equation by inserting the value in the equation and ~~seeing if the equality~~ is satisfied. We should always do this to check for any careless mistakes. For example, in 3a above we have two cases to consider. We call the left and ~~righthand~~ sides LHS and RHS respectively.	+	Keep in mind that we can always test our solution of an equation by inserting the solution's value into the equation and checking that it is satisfied. We should always do this to check for any careless mistakes. For example, in 3a above we have two cases to consider. We call the left-hand and right-hand sides LHS and RHS respectively.

	* <math>x = 2</math> gives that <math>\mbox{LHS } = 2^2 +2\cdot 2 - 8 = 4+4-8 = 0 = \mbox{RHS}</math>.		* <math>x = 2</math> gives that <math>\mbox{LHS } = 2^2 +2\cdot 2 - 8 = 4+4-8 = 0 = \mbox{RHS}</math>.
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	The functions		The functions
	{{Displayed math\|\|<math>\eqalign{y&=x^2-2x+5\cr y&=4-3x^2\cr y&=\textstyle\frac{1}{5}x^2 +3x}</math>}}		{{Displayed math\|\|<math>\eqalign{y&=x^2-2x+5\cr y&=4-3x^2\cr y&=\textstyle\frac{1}{5}x^2 +3x}</math>}}
-	are examples of functions ~~of the second degree~~. In general a function ~~of the second degree~~ can be written as	+	are examples of quadratic functions. In general a quadratic function can be written as
	{{Displayed math\|\|<math>y=ax^2+bx+c</math>}}		{{Displayed math\|\|<math>y=ax^2+bx+c</math>}}
	where <math>a</math>, <math>b</math>, <math>c</math> are constants and <math>a\ne0</math>.		where <math>a</math>, <math>b</math>, <math>c</math> are constants and <math>a\ne0</math>.

-	The graph for a function ~~of the second degree~~ is known as a parabola. The figures show the graphs of two typical parabolas <math>y=x^2</math> and <math>y=-x^2</math>.	+	The graph for a quadratic function is known as a parabola. The figures show the graphs of two typical parabolas <math>y=x^2</math> and <math>y=-x^2</math>.

	<center>{{:2.3 - Figure - The parabolas y = x² and y = -x²}}</center>		<center>{{:2.3 - Figure - The parabolas y = x² and y = -x²}}</center>

Eakc07 at 23:25, 11 November 2008

2008-11-11T23:25:46Z

←Older revision		Revision as of 23:25, 11 November 2008
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	== Parabolas ==		== Parabolas ==

-	~~Functions~~	+	The functions
	{{Displayed math\|\|<math>\eqalign{y&=x^2-2x+5\cr y&=4-3x^2\cr y&=\textstyle\frac{1}{5}x^2 +3x}</math>}}		{{Displayed math\|\|<math>\eqalign{y&=x^2-2x+5\cr y&=4-3x^2\cr y&=\textstyle\frac{1}{5}x^2 +3x}</math>}}
-	are examples of functions of the second degree. In general a function of the second degree can be written as	+	are examples of functions of the second degree. In general a function of the second degree can be written as
	{{Displayed math\|\|<math>y=ax^2+bx+c</math>}}		{{Displayed math\|\|<math>y=ax^2+bx+c</math>}}
	where <math>a</math>, <math>b</math>, <math>c</math> are constants and <math>a\ne0</math>.		where <math>a</math>, <math>b</math>, <math>c</math> are constants and <math>a\ne0</math>.
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	<center>{{:2.3 - Figure - The parabolas y = x² and y = -x²}}</center>		<center>{{:2.3 - Figure - The parabolas y = x² and y = -x²}}</center>
-	<center><small>The figure on the left shows the parabola <math>y=x^2</math> ~~and~~ figure to the right the parabola <math>y=-x^2</math>.</small></center>	+	<center><small>The figure on the left shows the parabola <math>y=x^2</math>. The figure to the right is the parabola <math>y=-x^2</math>.</small></center>


-	As the expression <math>x^2</math> is minimal when <math>x=0</math> the parabola <math>y=x^2</math> has a minimum when <math>x=0</math> ~~and~~ the parabola <math>y=-x^2</math> has a maximum when <math>x=0</math>.	+	As the expression <math>x^2</math> is minimal when <math>x=0</math>, the parabola <math>y=x^2</math> has a minimum when <math>x=0</math>. Similarly, the parabola <math>y=-x^2</math> has a maximum when <math>x=0</math>.

-	Note also that parabolas above are symmetrical about the <math>y</math>-axis as the value of <math>x^2</math> does not depend on the sign of <math>x</math>.	+	Note also that the parabolas above are symmetrical about the <math>y</math>-axis. This is because the value of <math>x^2</math> does not depend on the sign of <math>x</math>.

	<div class="exempel">		<div class="exempel">
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	\|\|<ol type="a">		\|\|<ol type="a">
	<li>Sketch the parabola <math>\ y=x^2-2</math>. <br><br>		<li>Sketch the parabola <math>\ y=x^2-2</math>. <br><br>
-	~~Compared~~ to the parabola <math>y=x^2</math> all points on the parabola (<math>y=x^2-2</math>) have <math>y</math>-values that are two units ~~smaller~~, so the parabola has been displaced downwards two units along the <math>y</math>-direction.</li>	+	Comparing it to the parabola <math>y=x^2</math>, we see that all points on the parabola (<math>y=x^2-2</math>) will have <math>y</math>-values that are two units less, so the parabola has been displaced downwards two units along the <math>y</math>-direction.</li>
	</ol>		</ol>
	\|align="right"\|{{:2.3 - Figure - The parabola y = x² - 2}}		\|align="right"\|{{:2.3 - Figure - The parabola y = x² - 2}}
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	\|\|<ol type="a" start=2>		\|\|<ol type="a" start=2>
	<li> Sketch the parabola <math>\ y=(x-2)^2</math>. <br><br>		<li> Sketch the parabola <math>\ y=(x-2)^2</math>. <br><br>
-	For the parabola <math>y=(x-2)^2</math> we need to choose <math>x</math>-values two units larger than for ~~the~~ parabola <math>y=x^2</math> to get the ~~corresponding~~ <math>y</math> ~~values~~. So the parabola <math>y=(x-2)^2</math> has been displaced two units to the right~~, compared to <math>y=x^2</math>~~.</li>	+	For the parabola <math>y=(x-2)^2</math>, we need to choose <math>x</math>-values that are two units larger than those for parabola <math>y=x^2</math> to get the same <math>y</math> value. So the parabola <math>y=(x-2)^2</math> has been displaced two units to the right.</li>
	</ol>		</ol>
	\|align="right"\|{{:2.3 - Figure - The parabola y = (x - 2)²}}		\|align="right"\|{{:2.3 - Figure - The parabola y = (x - 2)²}}
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	\|\|<ol type="a" start=3>		\|\|<ol type="a" start=3>
	<li> Sketch the parabola <math>\ y=2x^2</math>. <br><br>		<li> Sketch the parabola <math>\ y=2x^2</math>. <br><br>
-	Each point on the parabola <math>y=2x^2</math> has a <math>y</math>-value twice as large as the ~~corresponding~~ point with the same <math>x</math>-value on the parabola <math>y=x^2</math>. Thus parabola <math>y=2x^2</math> has been ~~increased~~ by a factor <math>2</math> in the <math>y</math>-direction ~~as compared~~ to <math>y=x^2</math>.	+	Each point on the parabola <math>y=2x^2</math> has a <math>y</math>-value twice as large as the point with the same <math>x</math>-value on the parabola <math>y=x^2</math>. Thus, the parabola <math>y=2x^2</math> has been stretched by a factor of <math>2</math> in the <math>y</math>-direction in comparison to <math>y=x^2</math>.
	</ol>		</ol>
	\|align="right"\|{{:2.3 - Figure - The parabola y = 2x²}}		\|align="right"\|{{:2.3 - Figure - The parabola y = 2x²}}
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	</div>		</div>

-	All sorts of parabolas can be handled by ~~the~~ completing the square ~~method~~.	+	All sorts of parabolas can be handled by completing the square.

	<div class="exempel">		<div class="exempel">
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	If one completes the square for the right-hand side		If one completes the square for the right-hand side
	{{Displayed math\|\|<math>x^2 +2x+2 = (x+1)^2 -1^2 +2 = (x+1)^2+1</math>}}		{{Displayed math\|\|<math>x^2 +2x+2 = (x+1)^2 -1^2 +2 = (x+1)^2+1</math>}}
-	we see from the resulting expression <math>y= (x+1)^2+1</math> that the parabola has been displaced one unit to the left along the <math>x</math>-direction ~~compared to~~ <math>y~~=x^2~~</math> (as ~~it stands~~ <math>(x~~+1)~~^2</math> ~~instead of <math>x^2</math>) and one unit upwards along the <math>y</math>-direction~~	+	we see from the resulting expression <math>y= (x+1)^2+1</math> that the parabola has been displaced one unit to the left along the <math>x</math>-direction and one unit up in the <math>y</math>-direction, as compared to <math>y=x^2</math>.
	\|\|{{:2.3 - Figure - The parabola y = x² + 2x + 2}}		\|\|{{:2.3 - Figure - The parabola y = x² + 2x + 2}}
	\|}		\|}
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	''' Example 7'''		''' Example 7'''

-	Determine where the parabola <math>\,y=x^2-4x+3\,</math> ~~cuts~~ the <math>x</math>-axis.	+	Determine where the parabola <math>\y=x^2-4x+3\</math> intersects the <math>x</math>-axis.


-	A point is on the <math>x</math>-axis if its <math>y</math>-coordinate is zero~~, and the~~ points on the parabola which have <math>y=0</math> have an <math>x</math>-coordinate that satisfies the equation	+	A point is on the <math>x</math>-axis if its <math>y</math>-coordinate is zero. The points on the parabola which have <math>y=0</math> have an <math>x</math>-coordinate that satisfies the equation
	{{Displayed math\|\|<math>x^2-4x+3=0\mbox{.}</math>}}		{{Displayed math\|\|<math>x^2-4x+3=0\mbox{.}</math>}}
-	Complete the square for the left-hand side,	+	Complete the square for the left-hand side
	{{Displayed math\|\|<math>x^2-4x+3=(x-2)^2-2^2+3=(x-2)^2-1</math>}}		{{Displayed math\|\|<math>x^2-4x+3=(x-2)^2-2^2+3=(x-2)^2-1</math>}}
	and this gives the equation		and this gives the equation
	{{Displayed math\|\|<math>(x-2)^2= 1 \; \mbox{.}</math>}}		{{Displayed math\|\|<math>(x-2)^2= 1 \; \mbox{.}</math>}}
-	After taking roots we get solutions	+	After taking roots we get the solutions
	*<math>x-2 =\sqrt{1} = 1,\quad</math> i.e. <math>\quad x=2+1=3</math>,		*<math>x-2 =\sqrt{1} = 1,\quad</math> i.e. <math>\quad x=2+1=3</math>,
	*<math>x-2 = -\sqrt{1} = -1,\quad</math> i.e. <math>\quad x=2-1=1</math>.		*<math>x-2 = -\sqrt{1} = -1,\quad</math> i.e. <math>\quad x=2-1=1</math>.
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	We complete the square		We complete the square
	{{Displayed math\|\|<math>x^2 +8x+19=(x+4)^2 -4^2 +19 = (x+4)^2 +3</math>}}		{{Displayed math\|\|<math>x^2 +8x+19=(x+4)^2 -4^2 +19 = (x+4)^2 +3</math>}}
-	and ~~then we~~ see that the ~~expression~~ must be ~~at least~~ equal to 3 because the square <math>(x+4)^2</math> is always greater than or equal to 0 regardless of what <math>x</math> is.	+	and see that the <math>y</math>-value must always be greater than or equal to 3. This is because the square <math>(x+4)^2</math> is always greater than or equal to 0 regardless of what <math>x</math> is.

-	In the figure below, we see that the whole parabola <math>y=x^2+8x+19</math> lies above the <math>x</math>-axis and has a minimum 3 at <math>x=-4</math>.	+	In the figure below we see that the whole parabola <math>y=x^2+8x+19</math> lies above the <math>x</math>-axis and has a minimum 3 at <math>x=-4</math>.

	<center>{{:2.3 - Figure - The parabola y = x² + 8x + 19}}</center>		<center>{{:2.3 - Figure - The parabola y = x² + 8x + 19}}</center>
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	'''Keep in mind that...'''		'''Keep in mind that...'''

-	You should devote a lot of time to doing algebra! Algebra is the alphabet of mathematics. Once you understand algebra~~, your~~ will enhance your understanding of statistics, areas, volumes and geometry.	+	You should devote a lot of time to doing algebra! Algebra is the alphabet of mathematics. Once you understand algebra you will enhance your understanding of statistics, areas, volumes and geometry.

Eakc07 at 22:42, 2 November 2008

2008-11-02T22:42:51Z

←Older revision		Revision as of 22:42, 2 November 2008
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	<li><math>x^2 = 4 \quad</math> has the roots <math>x=\sqrt{4} = 2</math> and <math>x=-\sqrt{4}= -2</math>.</li>		<li><math>x^2 = 4 \quad</math> has the roots <math>x=\sqrt{4} = 2</math> and <math>x=-\sqrt{4}= -2</math>.</li>
	<li><math>2x^2=18 \quad</math> is rewritten as <math>x^2=9</math> and has the roots <math>x=\sqrt9 = 3</math> and <math>x=-\sqrt9 = -3</math>.</li>		<li><math>2x^2=18 \quad</math> is rewritten as <math>x^2=9</math> and has the roots <math>x=\sqrt9 = 3</math> and <math>x=-\sqrt9 = -3</math>.</li>
-	<li><math>3x^2-15=0 \quad</math> can be rewritten as <math>x^2=5</math> and has the roots <math>x=\sqrt5 \approx 2\text{.}236</math> and <math>x=-\sqrt5 \approx -2{,}236</math>.</li>	+	<li><math>3x^2-15=0 \quad</math> can be rewritten as <math>x^2=5</math> and has the roots <math>x=\sqrt5 \approx 2\text{.}236</math> and <math>x=-\sqrt5 \approx -2\text{.}236</math>.</li>
-	<li><math>9x^2+25=0\quad</math> has no solutions because the left-hand side will always be greater than or equal to 25 regardless of the value of <math>x</math> (~~the~~ square <math>x^2</math> is always greater than or equal to zero).	+	<li><math>9x^2+25=0\quad</math> has no solutions because the left-hand side will always be greater than or equal to 25 regardless of the value of <math>x</math> (being a square, <math>x^2</math> is always greater than or equal to zero).
	</ol>		</ol>
	</div>		</div>
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	<li>Solve the equation <math>\ (x-1)^2 = 16</math>. <br><br>		<li>Solve the equation <math>\ (x-1)^2 = 16</math>. <br><br>
	By considering <math>x-1</math> as the unknown and taking the roots one finds the equation has two solutions		By considering <math>x-1</math> as the unknown and taking the roots one finds the equation has two solutions
-	*<math>x-1 =\sqrt{16} = 4\,</math> which gives ~~that~~ <math>x=1+4=5</math>,	+	*<math>x-1 =\sqrt{16} = 4\,</math> which gives <math>x=1+4=5</math>.
-	*<math>x-1 = -\sqrt{16} = -4\,</math> which gives ~~that~~ <math>x=1-4=-3</math>. </li>	+	*<math>x-1 = -\sqrt{16} = -4\,</math> which gives <math>x=1-4=-3</math>. </li>
	<li> Solve the equation <math>\ 2(x+1)^2 -8=0</math>. <br><br>		<li> Solve the equation <math>\ 2(x+1)^2 -8=0</math>. <br><br>
-	Move the term <math>8</math> over to the ~~right-hand~~ side and divide both sides by <math>2</math>,	+	Move the term <math>8</math> over to the righthand side and divide both sides by <math>2</math>,
	{{Displayed math\|\|<math>(x+1)^2=4 \; \mbox{.}</math>}}		{{Displayed math\|\|<math>(x+1)^2=4 \; \mbox{.}</math>}}
	Taking the roots gives:		Taking the roots gives:
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	</div>		</div>

-	To solve a quadratic equation generally, we use a technique called completing the square.	+	To solve a quadratic equation generally we use a technique called completing the square.

	If we consider the rule for expanding a quadratic,		If we consider the rule for expanding a quadratic,
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	One completes the square for <math>x^2+2x</math> (use <math>a=1</math> in the formula)		One completes the square for <math>x^2+2x</math> (use <math>a=1</math> in the formula)
	{{Displayed math\|\|<math>\underline{\vphantom{(}x^2+2x} -8 = \underline{(x+1)^2-1^2} -8 = (x+1)^2-9,</math>}}		{{Displayed math\|\|<math>\underline{\vphantom{(}x^2+2x} -8 = \underline{(x+1)^2-1^2} -8 = (x+1)^2-9,</math>}}
-	where the underlined terms are those involved in the completion of the square. ~~Thus the equation~~ can ~~be written as~~	+	where the underlined terms are those involved in the completion of the square. Using this we can write
	{{Displayed math\|\|<math>(x+1)^2 -9 = 0,</math>}}		{{Displayed math\|\|<math>(x+1)^2 -9 = 0,</math>}}
-	which we solve by taking roots	+	which we solve by taking roots, namely
-	*<math>x+1 =\sqrt{9} = 3\,</math> ~~and hence~~ <math>x=-1+3=2</math>,	+	*<math>x+1 =\sqrt{9} = 3\,</math> giving <math>x=-1+3=2</math>,
-	*<math>x+1 =-\sqrt{9} = -3\,</math> ~~and hence~~ <math>x=-1-3=-4</math>.</li>	+	*<math>x+1 =-\sqrt{9} = -3\,</math> giving <math>x=-1-3=-4</math>.</li>

	<li> Solve the equation <math>\ 2x^2 -2x - \frac{3}{2} = 0</math>. <br><br>		<li> Solve the equation <math>\ 2x^2 -2x - \frac{3}{2} = 0</math>. <br><br>
	Divide both sides by 2		Divide both sides by 2
	{{Displayed math\|\|<math>x^2-x-\textstyle\frac{3}{4}=0\mbox{.}</math>}}		{{Displayed math\|\|<math>x^2-x-\textstyle\frac{3}{4}=0\mbox{.}</math>}}
-	Complete the square of the ~~left-hand~~ side (use <math>a=-\tfrac{1}{2}</math>)	+	Complete the square of the lefthand side (use <math>a=-\tfrac{1}{2}</math>)
	{{Displayed math\|\|<math>\textstyle\underline{\vphantom{\bigl(\frac{3}{4}}x^2-x} -\frac{3}{4} = \underline{\bigl(x-\frac{1}{2}\bigr)^2 - \bigl(-\frac{1}{2}\bigr)^2} -\frac{3}{4}= \bigl(x-\frac{1}{2}\bigr)^2 -1</math>}}		{{Displayed math\|\|<math>\textstyle\underline{\vphantom{\bigl(\frac{3}{4}}x^2-x} -\frac{3}{4} = \underline{\bigl(x-\frac{1}{2}\bigr)^2 - \bigl(-\frac{1}{2}\bigr)^2} -\frac{3}{4}= \bigl(x-\frac{1}{2}\bigr)^2 -1</math>}}
-	~~and this~~ gives us the equation	+	This gives us the equation
	{{Displayed math\|\|<math>\textstyle\bigl(x-\frac{1}{2}\bigr)^2 - 1=0\mbox{.}</math>}}		{{Displayed math\|\|<math>\textstyle\bigl(x-\frac{1}{2}\bigr)^2 - 1=0\mbox{.}</math>}}
	Taking roots gives		Taking roots gives
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	'''Hint: '''		'''Hint: '''

-	Keep in mind that we can always test our solution to an equation by inserting the value in the equation and ~~see~~ if the ~~equation~~ is satisfied. We should always do this to check for any careless mistakes. For example, in 3a above, we have two cases to consider. We call the left- and ~~right-hand~~ sides LHS and RHS respectively:	+	Keep in mind that we can always test our solution to an equation by inserting the value in the equation and seeing if the equality is satisfied. We should always do this to check for any careless mistakes. For example, in 3a above we have two cases to consider. We call the left and righthand sides LHS and RHS respectively.

	* <math>x = 2</math> gives that <math>\mbox{LHS } = 2^2 +2\cdot 2 - 8 = 4+4-8 = 0 = \mbox{RHS}</math>.		* <math>x = 2</math> gives that <math>\mbox{LHS } = 2^2 +2\cdot 2 - 8 = 4+4-8 = 0 = \mbox{RHS}</math>.
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	<ol type="a">		<ol type="a">
	<li>Solve the equation <math>\ x^2-4x=0</math>. <br><br>		<li>Solve the equation <math>\ x^2-4x=0</math>. <br><br>
-	On the left-hand side, we can factor out an <math>x</math>	+	On the left-hand side we can factor out an <math>x</math>
	:<math>x(x-4)=0</math>.		:<math>x(x-4)=0</math>.
-	The equation on the ~~left-hand~~ side is zero when one of its factors is zero, which gives us two solutions	+	The equation on the lefthand side is zero when one of its factors is zero, which gives us two solutions
	*<math>x =0,\quad</math> or		*<math>x =0,\quad</math> or
	*<math>x-4=0\quad</math> which gives <math>\quad x=4</math>.</li>		*<math>x-4=0\quad</math> which gives <math>\quad x=4</math>.</li>
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	Functions		Functions
	{{Displayed math\|\|<math>\eqalign{y&=x^2-2x+5\cr y&=4-3x^2\cr y&=\textstyle\frac{1}{5}x^2 +3x}</math>}}		{{Displayed math\|\|<math>\eqalign{y&=x^2-2x+5\cr y&=4-3x^2\cr y&=\textstyle\frac{1}{5}x^2 +3x}</math>}}
-	are examples of functions of the second degree. In general, a function of the second degree can be written as	+	are examples of functions of the second degree. In general a function of the second degree can be written as
	{{Displayed math\|\|<math>y=ax^2+bx+c</math>}}		{{Displayed math\|\|<math>y=ax^2+bx+c</math>}}
-	where <math>a</math>, <math>b</math> ~~and~~ <math>c</math> are constants, and ~~where~~ <math>a\ne0</math>.	+	where <math>a</math>, <math>b</math>, <math>c</math> are constants and <math>a\ne0</math>.

-	The graph for a function of the second degree is known as a parabola ~~and the~~ figures show the graphs of two typical parabolas <math>y=x^2</math> and <math>y=-x^2</math>.	+	The graph for a function of the second degree is known as a parabola. The figures show the graphs of two typical parabolas <math>y=x^2</math> and <math>y=-x^2</math>.

	<center>{{:2.3 - Figure - The parabolas y = x² and y = -x²}}</center>		<center>{{:2.3 - Figure - The parabolas y = x² and y = -x²}}</center>
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	As the expression <math>x^2</math> is minimal when <math>x=0</math> the parabola <math>y=x^2</math> has a minimum when <math>x=0</math> and the parabola <math>y=-x^2</math> has a maximum when <math>x=0</math>.		As the expression <math>x^2</math> is minimal when <math>x=0</math> the parabola <math>y=x^2</math> has a minimum when <math>x=0</math> and the parabola <math>y=-x^2</math> has a maximum when <math>x=0</math>.

-	Note also that parabolas above are symmetrical about the <math>y</math>-axis, as the value of <math>x^2</math> does not depend on the sign of <math>x</math>.	+	Note also that parabolas above are symmetrical about the <math>y</math>-axis as the value of <math>x^2</math> does not depend on the sign of <math>x</math>.

	<div class="exempel">		<div class="exempel">
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	\|\|<ol type="a" start=3>		\|\|<ol type="a" start=3>
	<li> Sketch the parabola <math>\ y=2x^2</math>. <br><br>		<li> Sketch the parabola <math>\ y=2x^2</math>. <br><br>
-	Each point on the parabola <math>y=2x^2</math> has ~~twice as large~~ a <math>y</math>-value as the corresponding point with the same <math>x</math>-value on parabola <math>y=x^2</math>. Thus parabola <math>y=2x^2</math> has been increased by a factor <math>2</math> in the <math>y</math>-direction as compared to <math>y=x^2</math>.	+	Each point on the parabola <math>y=2x^2</math> has a <math>y</math>-value twice as large as the corresponding point with the same <math>x</math>-value on the parabola <math>y=x^2</math>. Thus parabola <math>y=2x^2</math> has been increased by a factor <math>2</math> in the <math>y</math>-direction as compared to <math>y=x^2</math>.
	</ol>		</ol>
	\|align="right"\|{{:2.3 - Figure - The parabola y = 2x²}}		\|align="right"\|{{:2.3 - Figure - The parabola y = 2x²}}
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	If one completes the square for the right-hand side		If one completes the square for the right-hand side
	{{Displayed math\|\|<math>x^2 +2x+2 = (x+1)^2 -1^2 +2 = (x+1)^2+1</math>}}		{{Displayed math\|\|<math>x^2 +2x+2 = (x+1)^2 -1^2 +2 = (x+1)^2+1</math>}}
-	we see from the resulting expression <math>y= (x+1)^2+1</math> that the parabola has been displaced one unit to the left along the <math>x</math>-direction, compared to <math>y=x^2</math> (as it stands <math>(x+1)^2</math> instead of <math>x^2</math>) and one unit upwards along the <math>y</math>-direction	+	we see from the resulting expression <math>y= (x+1)^2+1</math> that the parabola has been displaced one unit to the left along the <math>x</math>-direction compared to <math>y=x^2</math> (as it stands <math>(x+1)^2</math> instead of <math>x^2</math>) and one unit upwards along the <math>y</math>-direction
	\|\|{{:2.3 - Figure - The parabola y = x² + 2x + 2}}		\|\|{{:2.3 - Figure - The parabola y = x² + 2x + 2}}
	\|}		\|}

Eakc07 at 22:14, 2 November 2008

2008-11-02T22:14:52Z

←Older revision		Revision as of 22:14, 2 November 2008
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	<li><math>x^2 = 4 \quad</math> has the roots <math>x=\sqrt{4} = 2</math> and <math>x=-\sqrt{4}= -2</math>.</li>		<li><math>x^2 = 4 \quad</math> has the roots <math>x=\sqrt{4} = 2</math> and <math>x=-\sqrt{4}= -2</math>.</li>
	<li><math>2x^2=18 \quad</math> is rewritten as <math>x^2=9</math> and has the roots <math>x=\sqrt9 = 3</math> and <math>x=-\sqrt9 = -3</math>.</li>		<li><math>2x^2=18 \quad</math> is rewritten as <math>x^2=9</math> and has the roots <math>x=\sqrt9 = 3</math> and <math>x=-\sqrt9 = -3</math>.</li>
-	<li><math>3x^2-15=0 \quad</math> can be rewritten as <math>x^2=5</math> and has the roots <math>x=\sqrt5 \approx 2\~~rmtext~~{.}236</math> and <math>x=-\sqrt5 \approx -2{,}236</math>.</li>	+	<li><math>3x^2-15=0 \quad</math> can be rewritten as <math>x^2=5</math> and has the roots <math>x=\sqrt5 \approx 2\text{.}236</math> and <math>x=-\sqrt5 \approx -2{,}236</math>.</li>
	<li><math>9x^2+25=0\quad</math> has no solutions because the left-hand side will always be greater than or equal to 25 regardless of the value of <math>x</math> (the square <math>x^2</math> is always greater than or equal to zero).		<li><math>9x^2+25=0\quad</math> has no solutions because the left-hand side will always be greater than or equal to 25 regardless of the value of <math>x</math> (the square <math>x^2</math> is always greater than or equal to zero).
	</ol>		</ol>

Eakc07 at 22:14, 2 November 2008

2008-11-02T22:14:10Z

←Older revision		Revision as of 22:14, 2 November 2008
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	*Complete the square for expressions of degree two (second degree).		*Complete the square for expressions of degree two (second degree).
	*Solve quadratic equations by completing the square (not using a standard formula) and know how to check the answer.		*Solve quadratic equations by completing the square (not using a standard formula) and know how to check the answer.
-	*Factorise ~~expressions of the~~ second degree. (when possible).	+	*Factorise second degree expressions (when possible).
	*Directly solve factorised or almost factorised quadratic equations.		*Directly solve factorised or almost factorised quadratic equations.
	*Determine the minimum / maximum value of an expression of degree two.		*Determine the minimum / maximum value of an expression of degree two.
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	<ol type="a">		<ol type="a">
	<li><math>x^2 = 4 \quad</math> has the roots <math>x=\sqrt{4} = 2</math> and <math>x=-\sqrt{4}= -2</math>.</li>		<li><math>x^2 = 4 \quad</math> has the roots <math>x=\sqrt{4} = 2</math> and <math>x=-\sqrt{4}= -2</math>.</li>
-	<li><math>2x^2=18 \quad</math> is rewritten as <math>x^2=9</math> , and has the roots <math>x=\sqrt9 = 3</math> and <math>x=-\sqrt9 = -3</math>.</li>	+	<li><math>2x^2=18 \quad</math> is rewritten as <math>x^2=9</math> and has the roots <math>x=\sqrt9 = 3</math> and <math>x=-\sqrt9 = -3</math>.</li>
-	<li><math>3x^2-15=0 \quad</math> can be rewritten as <math>x^2=5</math> and has the roots <math>x=\sqrt5 \approx 2{,}236</math> and <math>x=-\sqrt5 \approx -2{,}236</math>.</li>	+	<li><math>3x^2-15=0 \quad</math> can be rewritten as <math>x^2=5</math> and has the roots <math>x=\sqrt5 \approx 2\rmtext{.}236</math> and <math>x=-\sqrt5 \approx -2{,}236</math>.</li>
	<li><math>9x^2+25=0\quad</math> has no solutions because the left-hand side will always be greater than or equal to 25 regardless of the value of <math>x</math> (the square <math>x^2</math> is always greater than or equal to zero).		<li><math>9x^2+25=0\quad</math> has no solutions because the left-hand side will always be greater than or equal to 25 regardless of the value of <math>x</math> (the square <math>x^2</math> is always greater than or equal to zero).
	</ol>		</ol>

Jdinglis at 14:00, 26 October 2008

2008-10-26T14:00:51Z

←Older revision		Revision as of 14:00, 26 October 2008
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-	'''Keep in mind that: '''	+	'''Keep in mind that...'''

-	~~Devote much~~ time to doing algebra! Algebra is the alphabet of mathematics. Once you understand algebra, your will enhance your understanding of statistics, areas, volumes and geometry.	+	You should devote a lot of time to doing algebra! Algebra is the alphabet of mathematics. Once you understand algebra, your will enhance your understanding of statistics, areas, volumes and geometry.

Tekbot: Robot: Automated text replacement (-{{:2.3 - Figur - Parablerna y = x² och y = -x²}} +{{:2.3 - Figure - The parabolas y = x² and y = -x²}})

2008-09-12T12:52:38Z

Robot: Automated text replacement (-{{:2.3 - Figur - Parablerna y = x² och y = -x²}} +{{:2.3 - Figure - The parabolas y = x² and y = -x²}})

←Older revision		Revision as of 12:52, 12 September 2008
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	The graph for a function of the second degree is known as a parabola and the figures show the graphs of two typical parabolas <math>y=x^2</math> and <math>y=-x^2</math>.		The graph for a function of the second degree is known as a parabola and the figures show the graphs of two typical parabolas <math>y=x^2</math> and <math>y=-x^2</math>.

-	<center>{{:2.3 - ~~Figur~~ - ~~Parablerna~~ y = x² ~~och~~ y = -x²}}</center>	+	<center>{{:2.3 - Figure - The parabolas y = x² and y = -x²}}</center>
	<center><small>The figure on the left shows the parabola <math>y=x^2</math> and figure to the right the parabola <math>y=-x^2</math>.</small></center>		<center><small>The figure on the left shows the parabola <math>y=x^2</math> and figure to the right the parabola <math>y=-x^2</math>.</small></center>

Tekbot: Robot: Automated text replacement (-{{:2.3 - Figur - Parabeln y = x² - 4x + 3}} +{{:2.3 - Figure - The parabola y = x² - 4x + 3}})

2008-09-12T12:51:32Z

Robot: Automated text replacement (-{{:2.3 - Figur - Parabeln y = x² - 4x + 3}} +{{:2.3 - Figure - The parabola y = x² - 4x + 3}})

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	The parabola cuts the <math>x</math>-axis in points <math>(1,0)</math> and <math>(3,0)</math>.		The parabola cuts the <math>x</math>-axis in points <math>(1,0)</math> and <math>(3,0)</math>.

-	<center>{{:2.3 - ~~Figur~~ - ~~Parabeln~~ y = x² - 4x + 3}}</center>	+	<center>{{:2.3 - Figure - The parabola y = x² - 4x + 3}}</center>

	</div>		</div>

Tekbot: Robot: Automated text replacement (-{{:2.3 - Figur - Parabeln y = x² - 2}} +{{:2.3 - Figure - The parabola y = x² - 2}})

2008-09-12T12:50:27Z

Robot: Automated text replacement (-{{:2.3 - Figur - Parabeln y = x² - 2}} +{{:2.3 - Figure - The parabola y = x² - 2}})

←Older revision		Revision as of 12:50, 12 September 2008
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	Compared to the parabola <math>y=x^2</math> all points on the parabola (<math>y=x^2-2</math>) have <math>y</math>-values that are two units smaller, so the parabola has been displaced downwards two units along the <math>y</math>-direction.</li>		Compared to the parabola <math>y=x^2</math> all points on the parabola (<math>y=x^2-2</math>) have <math>y</math>-values that are two units smaller, so the parabola has been displaced downwards two units along the <math>y</math>-direction.</li>
	</ol>		</ol>
-	\|align="right"\|{{:2.3 - ~~Figur~~ - ~~Parabeln~~ y = x² - 2}}	+	\|align="right"\|{{:2.3 - Figure - The parabola y = x² - 2}}
	\|}		\|}

Tekbot: Robot: Automated text replacement (-{{:2.3 - Figur - Parabeln y = x² + 8x + 19}} +{{:2.3 - Figure - The parabola y = x² + 8x + 19}})

2008-09-12T12:49:21Z

Robot: Automated text replacement (-{{:2.3 - Figur - Parabeln y = x² + 8x + 19}} +{{:2.3 - Figure - The parabola y = x² + 8x + 19}})

←Older revision		Revision as of 12:49, 12 September 2008
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	In the figure below, we see that the whole parabola <math>y=x^2+8x+19</math> lies above the <math>x</math>-axis and has a minimum 3 at <math>x=-4</math>.		In the figure below, we see that the whole parabola <math>y=x^2+8x+19</math> lies above the <math>x</math>-axis and has a minimum 3 at <math>x=-4</math>.

-	<center>{{:2.3 - ~~Figur~~ - ~~Parabeln~~ y = x² + 8x + 19}}</center>	+	<center>{{:2.3 - Figure - The parabola y = x² + 8x + 19}}</center>
	</div>		</div>