2.3 Quadratic expressions
From Förberedande kurs i matematik 1
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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{ | + | <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> ( | + | <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 | + | *<math>x-1 =\sqrt{16} = 4\,</math> which gives <math>x=1+4=5</math>. |
- | *<math>x-1 = -\sqrt{16} = -4\,</math> which gives | + | *<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 | + | 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 | + | 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. | + | 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> | + | *<math>x+1 =\sqrt{9} = 3\,</math> giving <math>x=-1+3=2</math>, |
- | *<math>x+1 =-\sqrt{9} = -3\,</math> | + | *<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 | + | 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>}} | ||
- | + | 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 | + | 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 | + | 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 | + | 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 | + | 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> | + | 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 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 | + | 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 | + | 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 | + | 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}} | ||
|} | |} |
Revision as of 22:42, 2 November 2008
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
- \displaystyle x^2 = 4 \quad has the roots \displaystyle x=\sqrt{4} = 2 and \displaystyle x=-\sqrt{4}= -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.
- \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.
- \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
- 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.
- 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
- 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.
- 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
- 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.
- \displaystyle x =0,\quad or
- \displaystyle x-4=0\quad which gives \displaystyle \quad x=4.
Parabolas
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.
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 and the parabola \displaystyle y=-x^2 has a maximum when \displaystyle x=0.
Note also that parabolas above are symmetrical about the \displaystyle y-axis as the value of \displaystyle x^2 does not depend on the sign of \displaystyle x.
Example 5
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All sorts of parabolas can be handled by the completing the square method.
Example 6
Sketch the parabola \displaystyle \ y=x^2+2x+2.
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 compared to \displaystyle y=x^2 (as it stands \displaystyle (x+1)^2 instead of \displaystyle x^2) and one unit upwards along the \displaystyle y-direction |
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Example 7
Determine where the parabola \displaystyle \,y=x^2-4x+3\, cuts the \displaystyle x-axis.
A point is on the \displaystyle x-axis if its \displaystyle y-coordinate is zero, and 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 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).
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 then we see that the expression must be at least equal to 3 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.
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, 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
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