2.3 Quadratic expressions

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{{Vald flik|[[2.3 Andragradsuttryck|Theory]]}}
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{{Selected tab|[[2.3 Quadratic expressions|Theory]]}}
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{{Ej vald flik|[[2.3 Exercises|Exercises]]}}
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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.
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*Factorise expressions of the second degree. (when possible).
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*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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A quadratic equation is one that can be written as
A quadratic equation is one that can be written as
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{{Fristående formel||<math>x^2+px+q=0</math>}}
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{{Displayed math||<math>x^2+px+q=0</math>}}
where <math>x</math> is the unknown and <math>p</math> and <math>q</math> are constants.
where <math>x</math> is the unknown and <math>p</math> and <math>q</math> are constants.
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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>
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<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>
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<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\text{.}236</math> and <math>x=-\sqrt5 \approx -2\text{.}236</math>.</li>
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<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).
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<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
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*<math>x-1 =\sqrt{16} = 4\,</math> which gives that <math>x=1+4=5</math>,
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*<math>x-1 =\sqrt{16} = 4\,</math> which gives <math>x=1+4=5</math>.
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*<math>x-1 = -\sqrt{16} = -4\,</math> which gives that <math>x=1-4=-3</math>. </li>
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*<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>
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Move the term <math>8</math> over to the right-hand side and divide both sides by <math>2</math>,
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Move the term <math>8</math> over to the righthand side and divide both sides by <math>2</math>,
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{{Fristående formel||<math>(x+1)^2=4 \; \mbox{.}</math>}}
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{{Displayed math||<math>(x+1)^2=4 \; \mbox{.}</math>}}
Taking the roots gives:
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=1\,\mbox{,}</math>
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</div>
</div>
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To solve a quadratic equation generally, we use a technique called completing the square.
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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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{{Fristående formel||<math>x^2 + 2ax + a^2 = (x+a)^2</math>}}
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{{Displayed math||<math>x^2 + 2ax + a^2 = (x+a)^2</math>}}
and subtract the <math>a^2</math> from both sides we get
and subtract the <math>a^2</math> from both sides we get
<div class="regel">
<div class="regel">
'''Completing the square:'''
'''Completing the square:'''
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{{Fristående formel||<math>x^2 +2ax = (x+a)^2 -a^2</math>}}
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{{Displayed math||<math>x^2 +2ax = (x+a)^2 -a^2</math>}}
</div>
</div>
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<li> Solve the equation <math>\ x^2 +2x -8=0</math>. <br><br>
<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)
One completes the square for <math>x^2+2x</math> (use <math>a=1</math> in the formula)
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{{Fristående formel||<math>\underline{\vphantom{(}x^2+2x} -8 = \underline{(x+1)^2-1^2} -8 = (x+1)^2-9,</math>}}
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{{Displayed math||<math>\underline{\vphantom{(}x^2+2x} -8 = \underline{(x+1)^2-1^2} -8 = (x+1)^2-9,</math>}}
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where the underlined terms are those involved in the completion of the square. Thus the equation can be written as
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where the underlined terms are those involved in the completion of the square. Using this we can write
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{{Fristående formel||<math>(x+1)^2 -9 = 0,</math>}}
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{{Displayed math||<math>(x+1)^2 -9 = 0,</math>}}
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which we solve by taking roots
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which we solve by taking roots, namely
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*<math>x+1 =\sqrt{9} = 3\,</math> and hence <math>x=-1+3=2</math>,
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*<math>x+1 =\sqrt{9} = 3\,</math> giving <math>x=-1+3=2</math>,
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*<math>x+1 =-\sqrt{9} = -3\,</math> and hence <math>x=-1-3=-4</math>.</li>
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*<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
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{{Fristående formel||<math>x^2-x-\textstyle\frac{3}{4}=0\mbox{.}</math>}}
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{{Displayed math||<math>x^2-x-\textstyle\frac{3}{4}=0\mbox{.}</math>}}
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Complete the square of the left-hand side (use <math>a=-\tfrac{1}{2}</math>)
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Complete the square of the lefthand side (use <math>a=-\tfrac{1}{2}</math>)
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{{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>}}
+
{{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>}}
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and this gives us the equation
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This gives us the equation
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{{Fristående formel||<math>\textstyle\bigl(x-\frac{1}{2}\bigr)^2 - 1=0\mbox{.}</math>}}
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{{Displayed math||<math>\textstyle\bigl(x-\frac{1}{2}\bigr)^2 - 1=0\mbox{.}</math>}}
Taking roots gives
Taking roots gives
*<math>x-\tfrac{1}{2} =\sqrt{1} = 1, \quad</math> i.e. <math>\quad x=\tfrac{1}{2}+1=\tfrac{3}{2}</math>,
*<math>x-\tfrac{1}{2} =\sqrt{1} = 1, \quad</math> i.e. <math>\quad x=\tfrac{1}{2}+1=\tfrac{3}{2}</math>,
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'''Hint: '''
'''Hint: '''
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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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Using the completing the square method it is possible to show that the general quadratic equation
Using the completing the square method it is possible to show that the general quadratic equation
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{{Fristående formel||<math>x^2+px+q=0</math>}}
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{{Displayed math||<math>x^2+px+q=0</math>}}
has the solutions
has the solutions
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{{Fristående formel||<math>x = - \displaystyle\frac{p}{2} \pm \sqrt{\left(\frac{p}{2}\right)^2-q}</math>}}
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{{Displayed math||<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.
provided that the term inside the root sign is not negative.
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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>
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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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== Parabolas ==
== Parabolas ==
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Functions
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The functions
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{{Fristående formel||<math>\eqalign{y&=x^2-2x+5\cr y&=4-3x^2\cr y&=\textstyle\frac{1}{5}x^2 +3x}</math>}}
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{{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
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{{Fristående formel||<math>y=ax^2+bx+c</math>}}
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{{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 - Figur - Parablerna y = x² och y = -x²}}</center>
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<center>{{:2.3 - Figure - The parabolas y = x² and y = -x²}}</center>
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<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>
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<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>
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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>. Similarly, the parabola <math>y=-x^2</math> has a maximum when <math>x=0</math>.
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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>
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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>
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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>
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|align="right"|{{:2.3 - Figur - Parabeln y = x² - 2}}
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|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>
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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>
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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>
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|align="right"|{{:2.3 - Figur - Parabeln y = (x - 2)²}}
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|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>
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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>.
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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>
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|align="right"|{{:2.3 - Figur - Parabeln y = 2x²}}
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|align="right"|{{:2.3 - Figure - The parabola y = 2x²}}
|}
|}
</div>
</div>
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All sorts of parabolas can be handled by the completing the square method.
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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
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{{Fristående formel||<math>x^2 +2x+2 = (x+1)^2 -1^2 +2 = (x+1)^2+1</math>}}
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{{Displayed math||<math>x^2 +2x+2 = (x+1)^2 -1^2 +2 = (x+1)^2+1</math>}}
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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>.
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||{{:2.3 - Figur - Parabeln y = x² + 2x + 2}}
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||{{:2.3 - Figure - The parabola y = x² + 2x + 2}}
|}
|}
</div>
</div>
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''' Example 7'''
''' Example 7'''
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Determine where the parabola <math>\,y=x^2-4x+3\,</math> cuts the <math>x</math>-axis.
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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
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{{Fristående formel||<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
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{{Fristående formel||<math>x^2-4x+3=(x-2)^2-2^2+3=(x-2)^2-1</math>}}
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{{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
-
{{Fristående formel||<math>(x-2)^2= 1 \; \mbox{.}</math>}}
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{{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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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>.
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<center>{{:2.3 - Figur - Parabeln y = x² - 4x + 3}}</center>
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<center>{{:2.3 - Figure - The parabola y = x² - 4x + 3}}</center>
</div>
</div>
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We complete the square
We complete the square
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{{Fristående formel||<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>.
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<center>{{:2.3 - Figur - Parabeln y = x² + 8x + 19}}</center>
+
<center>{{:2.3 - Figure - The parabola y = x² + 8x + 19}}</center>
</div>
</div>
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'''Keep in mind that: '''
+
'''Keep in mind that...'''
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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 you will enhance your understanding of statistics, areas, volumes and geometry.

Current revision

       Theory          Exercises      

Contents:

  • Completing the square method
  • Quadratic equations
  • Factorising
  • Parabolas

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 second degree expressions (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

\displaystyle x^2+px+q=0

where \displaystyle x is the unknown and \displaystyle p and \displaystyle q are constants.


Simpler forms of quadratic equations can be solved directly by taking roots.

The equation \displaystyle x^2=a where \displaystyle a is a positive number has two solutions (roots) \displaystyle x=\sqrt{a} and \displaystyle x=-\sqrt{a}.

Example 1

  1. \displaystyle x^2 = 4 \quad has the roots \displaystyle x=\sqrt{4} = 2 and \displaystyle x=-\sqrt{4}= -2.
  2. \displaystyle 2x^2=18 \quad is rewritten as \displaystyle x^2=9 and has the roots \displaystyle x=\sqrt9 = 3 and \displaystyle x=-\sqrt9 = -3.
  3. \displaystyle 3x^2-15=0 \quad can be rewritten as \displaystyle x^2=5 and has the roots \displaystyle x=\sqrt5 \approx 2\text{.}236 and \displaystyle x=-\sqrt5 \approx -2\text{.}236.
  4. \displaystyle 9x^2+25=0\quad has no solutions because the left-hand side will always be greater than or equal to 25 regardless of the value of \displaystyle x (being a square, \displaystyle x^2 is always greater than or equal to zero).

Example 2

  1. Solve the equation \displaystyle \ (x-1)^2 = 16.

    By considering \displaystyle x-1 as the unknown and taking the roots one finds the equation has two solutions
    • \displaystyle x-1 =\sqrt{16} = 4\, which gives \displaystyle x=1+4=5.
    • \displaystyle x-1 = -\sqrt{16} = -4\, which gives \displaystyle x=1-4=-3.
  2. Solve the equation \displaystyle \ 2(x+1)^2 -8=0.

    Move the term \displaystyle 8 over to the righthand side and divide both sides by \displaystyle 2,
    \displaystyle (x+1)^2=4 \; \mbox{.}

    Taking the roots gives:

    • \displaystyle x+1 =\sqrt{4} = 2, \quad \mbox{dvs.} \quad x=-1+2=1\,\mbox{,}
    • \displaystyle x+1 = -\sqrt{4} = -2, \quad \mbox{dvs.} \quad x=-1-2=-3\,\mbox{.}

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

If we consider the rule for expanding a quadratic,

\displaystyle x^2 + 2ax + a^2 = (x+a)^2

and subtract the \displaystyle a^2 from both sides we get

Completing the square:

\displaystyle x^2 +2ax = (x+a)^2 -a^2

Example 3

  1. Solve the equation \displaystyle \ x^2 +2x -8=0.

    One completes the square for \displaystyle x^2+2x (use \displaystyle a=1 in the formula)
    \displaystyle \underline{\vphantom{(}x^2+2x} -8 = \underline{(x+1)^2-1^2} -8 = (x+1)^2-9,

    where the underlined terms are those involved in the completion of the square. Using this we can write

    \displaystyle (x+1)^2 -9 = 0,

    which we solve by taking roots, namely

    • \displaystyle x+1 =\sqrt{9} = 3\, giving \displaystyle x=-1+3=2,
    • \displaystyle x+1 =-\sqrt{9} = -3\, giving \displaystyle x=-1-3=-4.
  2. Solve the equation \displaystyle \ 2x^2 -2x - \frac{3}{2} = 0.

    Divide both sides by 2
    \displaystyle x^2-x-\textstyle\frac{3}{4}=0\mbox{.}

    Complete the square of the lefthand side (use \displaystyle a=-\tfrac{1}{2})

    \displaystyle \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

    This gives us the equation

    \displaystyle \textstyle\bigl(x-\frac{1}{2}\bigr)^2 - 1=0\mbox{.}

    Taking roots gives

    • \displaystyle x-\tfrac{1}{2} =\sqrt{1} = 1, \quad i.e. \displaystyle \quad x=\tfrac{1}{2}+1=\tfrac{3}{2},
    • \displaystyle x-\tfrac{1}{2}= -\sqrt{1} = -1, \quad i.e. \displaystyle \quad x=\tfrac{1}{2}-1= -\tfrac{1}{2}.

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.

  • \displaystyle x = 2 gives that \displaystyle \mbox{LHS } = 2^2 +2\cdot 2 - 8 = 4+4-8 = 0 = \mbox{RHS}.
  • \displaystyle x = -4 gives that \displaystyle \mbox{LHS } = (-4)^2 + 2\cdot(-4) -8 = 16-8-8 = 0 = \mbox{RHS}.

In both cases we arrive at LHS = RHS. The equation is satisfied in both cases.

Using the completing the square method it is possible to show that the general quadratic equation

\displaystyle x^2+px+q=0

has the solutions

\displaystyle x = - \displaystyle\frac{p}{2} \pm \sqrt{\left(\frac{p}{2}\right)^2-q}

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.

Example 4

  1. Solve the equation \displaystyle \ x^2-4x=0.

    On the left-hand side we can factor out an \displaystyle x
    \displaystyle x(x-4)=0.
    The equation on the lefthand side is zero when one of its factors is zero, which gives us two solutions
    • \displaystyle x =0,\quad or
    • \displaystyle x-4=0\quad which gives \displaystyle \quad x=4.


Parabolas

The functions

\displaystyle \eqalign{y&=x^2-2x+5\cr y&=4-3x^2\cr y&=\textstyle\frac{1}{5}x^2 +3x}

are examples of functions of the second degree. In general a function of the second degree can be written as

\displaystyle y=ax^2+bx+c

where \displaystyle a, \displaystyle b, \displaystyle c are constants and \displaystyle a\ne0.

The graph for a function of the second degree is known as a parabola. The figures show the graphs of two typical parabolas \displaystyle y=x^2 and \displaystyle y=-x^2.

[Image]

The figure on the left shows the parabola \displaystyle y=x^2. The figure to the right is the parabola \displaystyle y=-x^2.


As the expression \displaystyle x^2 is minimal when \displaystyle x=0, the parabola \displaystyle y=x^2 has a minimum when \displaystyle x=0. Similarly, the parabola \displaystyle y=-x^2 has a maximum when \displaystyle x=0.

Note also that the parabolas above are symmetrical about the \displaystyle y-axis. This is because the value of \displaystyle x^2 does not depend on the sign of \displaystyle x.

Example 5

  1. Sketch the parabola \displaystyle \ y=x^2-2.

    Comparing it to the parabola \displaystyle y=x^2, we see that all points on the parabola (\displaystyle y=x^2-2) will have \displaystyle y-values that are two units less, so the parabola has been displaced downwards two units along the \displaystyle y-direction.

[Image]

  1. Sketch the parabola \displaystyle \ y=(x-2)^2.

    For the parabola \displaystyle y=(x-2)^2, we need to choose \displaystyle x-values that are two units larger than those for parabola \displaystyle y=x^2 to get the same \displaystyle y value. So the parabola \displaystyle y=(x-2)^2 has been displaced two units to the right.

[Image]

  1. Sketch the parabola \displaystyle \ y=2x^2.

    Each point on the parabola \displaystyle y=2x^2 has a \displaystyle y-value twice as large as the point with the same \displaystyle x-value on the parabola \displaystyle y=x^2. Thus, the parabola \displaystyle y=2x^2 has been stretched by a factor of \displaystyle 2 in the \displaystyle y-direction in comparison to \displaystyle y=x^2.

[Image]

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

Example 6

Sketch the parabola \displaystyle \ y=x^2+2x+2.


If one completes the square for the right-hand side

\displaystyle x^2 +2x+2 = (x+1)^2 -1^2 +2 = (x+1)^2+1

we see from the resulting expression \displaystyle y= (x+1)^2+1 that the parabola has been displaced one unit to the left along the \displaystyle x-direction and one unit up in the \displaystyle y-direction, as compared to \displaystyle y=x^2.

[Image]

Example 7

Determine where the parabola \displaystyle \y=x^2-4x+3\ intersects the \displaystyle x-axis.


A point is on the \displaystyle x-axis if its \displaystyle y-coordinate is zero. The points on the parabola which have \displaystyle y=0 have an \displaystyle x-coordinate that satisfies the equation

\displaystyle x^2-4x+3=0\mbox{.}

Complete the square for the left-hand side

\displaystyle x^2-4x+3=(x-2)^2-2^2+3=(x-2)^2-1

and this gives the equation

\displaystyle (x-2)^2= 1 \; \mbox{.}

After taking roots we get the solutions

  • \displaystyle x-2 =\sqrt{1} = 1,\quad i.e. \displaystyle \quad x=2+1=3,
  • \displaystyle x-2 = -\sqrt{1} = -1,\quad i.e. \displaystyle \quad x=2-1=1.

The parabola cuts the \displaystyle x-axis in points \displaystyle (1,0) and \displaystyle (3,0).

[Image]

Example 8

Determine the minimum value of the expression \displaystyle \,x^2+8x+19\,.


We complete the square

\displaystyle x^2 +8x+19=(x+4)^2 -4^2 +19 = (x+4)^2 +3

and see that the \displaystyle y-value must always be greater than or equal to 3. This is because the square \displaystyle (x+4)^2 is always greater than or equal to 0 regardless of what \displaystyle x is.

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

[Image]


Exercises

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...

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.


Reviews

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

Learn more about quadratic equations in the English Wikipedia

Learn more about quadratic equations in mathworld

101 uses of a quadratic equation - by Chris Budd and Chris Sangwin


Useful web sites

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