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From Förberedande kurs i matematik 1

for 2.3 Quadratic expressions
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__NOTOC__
{| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"
| style="border-bottom:1px solid #797979" width="5px" | &nbsp;
{{Vald flik|[[2.3 Andragradsuttryck|Theory]]}}
{{Ej vald flik|[[2.3 Övningar|Exercises]]}}
| style="border-bottom:1px solid #797979"  width="100%"| &nbsp;
|}

{{Info|
'''Content:'''
*Completing the square method 
*Quadratic equations 
* Faktorising
* Parabolas
}}

{{Info|
'''Learning outcomes:'''

After this section, you will have learned to:
*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. 
*Factorise expressions of the second degree. (when possible).
*Directly solve factorised or almost factorised quadratic equations. 
*Determine the minimum / maximum value of an expression of degree two. 
*Sketch  parabolas by completing the square method. 
}}

== quadratic equations  ==

A quadratic equation is one that can be written as 
{{Fristående formel||<math>x^2+px+q=0</math>}}
where <math>x</math> is the unknown and <math>p</math> and <math>q</math> are constants. 


Simpler forms of quadratic equations can be solved directly by taking roots. 
<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>.
</div>

<div class="exempel">
''' Example 1'''

<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>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>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>
</div>

<div class="exempel">
''' Example 2'''

<ol type="a">
  <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
*<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 that <math>x=1-4=-3</math>. </li>
  <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>, 
{{Fristående formel||<math>(x+1)^2=4 \; \mbox{.}</math>}}
Taking the roots gives: 
*<math>x+1 =\sqrt{4} = 2, \quad \mbox{dvs.} \quad x=-1+2=1\,\mbox{,}</math> 
*<math>x+1 = -\sqrt{4} = -2, \quad \mbox{dvs.} \quad x=-1-2=-3\,\mbox{.}</math></li>
</ol>
</div>

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

If we consider the rule for expanding a quadratic, 
{{Fristående formel||<math>x^2 + 2ax + a^2 = (x+a)^2</math>}}
and subtract the <math>a^2</math> from both sides we get 

<div class="regel">
'''Kvadratkomplettering:'''
{{Fristående formel||<math>x^2 +2ax = (x+a)^2 -a^2</math>}}
</div>

<div class="exempel">
''' Example 3'''

<ol type="a">
  <li> Solve the equation <math>\ x^2 +2x -8=0</math>. <br><br>
One completes the square  for  <math>x^2+2x</math> (use  <math>a=1</math> in the formula)
{{Fristående formel||<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 
{{Fristående formel||<math>(x+1)^2 -9 = 0,</math>}}
which we solve by taking roots 
*<math>x+1 =\sqrt{9} = 3\,</math> and hence <math>x=-1+3=2</math>,
*<math>x+1 =-\sqrt{9} = -3\,</math> and hence <math>x=-1-3=-4</math>.</li>

  <li> Solve the equation <math>\ 2x^2 -2x - \frac{3}{2} = 0</math>. <br><br>
Divide both sides by 2 
{{Fristående formel||<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>)
{{Fristående formel||<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
{{Fristående formel||<math>\textstyle\bigl(x-\frac{1}{2}\bigr)^2 - 1=0\mbox{.}</math>}}
Taking roots gives
*<math>x-\tfrac{1}{2} =\sqrt{1} = 1, \quad</math> dvs. <math>\quad x=\tfrac{1}{2}+1=\tfrac{3}{2}</math>, 
*<math>x-\tfrac{1}{2}= -\sqrt{1} = -1, \quad</math> dvs. <math>\quad x=\tfrac{1}{2}-1= -\tfrac{1}{2}</math>.</li>
</ol>
</div>

<div class="tips">
'''Hint: '''

Keep in mind that we can always test solutions 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  for LH and RH respectively: 

* <math>x = 2</math> gives att <math>\mbox{LH } = 2^2 +2\cdot 2 - 8 = 4+4-8 = 0 = \mbox{RH}</math>.

* <math>x = -4</math>  gives att <math>\mbox{LH } = (-4)^2 + 2\cdot(-4) -8 = 16-8-8 = 0 = \mbox{RH}</math>.

In both cases we arrive at  LH = RH. The equation is satisfied in both cases. 
</div>

Using the completing the square method it is possible to show that the general quadratic equation
{{Fristående formel||<math>x^2+px+q=0</math>}}
has the solutions
{{Fristående formel||<math>x = - \displaystyle\frac{p}{2} \pm \sqrt{\left(\frac{p}{2}\right)^2-q}</math>}} 
provided that the term inside  the root sign  is not negative. 

Sometimes one can  factorise the equations directly and thus immediately see what the solutions are. 

<div class="exempel">
''' Example 4'''

<ol type="a">
  <li>Solve the equation  <math>\ x^2-4x=0</math>. <br><br>
On the left-hand side, we can factorise out an <math>x</math>
:<math>x(x-4)=0</math>.
The equations left-hand side is zero when one of its factors is zero, which gives us two solutions 
*<math>x =0,\quad</math> or
*<math>x-4=0\quad</math> or. <math>\quad x=4</math>.</li>
</ol>
</div>


== Parabolas ==

Functions
{{Fristående formel||<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 
{{Fristående formel||<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>. 

The graph for a functions 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><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>


As the expression <math>x^2</math> is minimum 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>. 

<div class="exempel">
''' Example 5'''

{| width="100%"
||<ol type="a">
  <li>Sketch the parabola <math>\ y=x^2-2</math>. <br><br>
Compared to 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>
|align="right"|{{:2.3 - Figur - Parabeln y = x² - 2}}
|}

{| width="100%"
||<ol type="a" start=2>
  <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>
</ol>
|align="right"|{{:2.3 - Figur - Parabeln y = (x - 2)²}}
|}

{| width="100%"
||<ol type="a" start=3>
  <li> Sketch the  parabola  <math>\ y=2x^2</math>. <br><br>
 Each point on the parabola <math>y=2x^2</math> has twice a large   <math>y</math>-value than 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>. 
</ol>
|align="right"|{{:2.3 - Figur - Parabeln y = 2x²}}
|}
</div>

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

<div class="exempel">
'''Exempel 6'''

{| width="100%"
||Sketch the  parabola <math>\ y=x^2+2x+2</math>. 


If  one  completes the square for the right-hand side 
{{Fristående formel||<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
||{{:2.3 - Figur - Parabeln y = x² + 2x + 2}}
|}
</div>

<div class="exempel">
''' Example 7'''

Determine where the parabola  <math>\,y=x^2-4x+3\,</math> cuts <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 
{{Fristående formel||<math>x^2-4x+3=0\mbox{.}</math>}}
Complete the square for the left-hand side,
{{Fristående formel||<math>x^2-4x+3=(x-2)^2-2^2+3=(x-2)^2-1</math>}}
and this gives the equation 
{{Fristående formel||<math>(x-2)^2= 1 \; \mbox{.}</math>}}
After taking roots we get solutions 
*<math>x-2 =\sqrt{1} = 1,\quad</math> dvs. <math>\quad x=2+1=3</math>, 
*<math>x-2 = -\sqrt{1} = -1,\quad</math> dvs. <math>\quad x=2-1=1</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>

</div>

<div class="exempel">
''' Example 8'''

Determine the minimum value of the expression  <math>\,x^2+8x+19\,</math> antar.


We completes the square 
{{Fristående formel||<math>x^2 +8x+19=(x+4)^2 -4^2 +19 = (x+4)^2 +3</math>}}
and then we see that the term 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. 

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> 
</div>


[[2.3 Övningar|Exercises]]

<div class="inforuta" style="width:580px;">
'''Study advice'''

'''Basic and final tests'''

After you have read the text and worked through the exercises, you should do the basic and final tests to pass this section. You can find the link to the tests in your student lounge. 


'''Keep in mind that: '''

Devote much  time to doing algebra! Algebra is mathematic' s alphabet. Once you understand algebra, your will enhance your  understanding of statistics, areas, volumes and geometry. 


'''Reviews'''

For those of you who want to deepen your studies or need more detailed explanations consider the following references 

[http://en.wikipedia.org/wiki/Quadratic_equation   Learn more about quadratic equations in the English Wikipedia ]

[http://mathworld.wolfram.com/QuadraticEquation.html   Learn more about quadratic equations in mathworld ]

[http://plus.maths.org/issue29/features/quadratic/index-gifd.html   101 uses of a quadratic equation - by Chris Budd and Chris Sangwin]


'''Länktips'''
</div>
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