2.3 Quadratische Gleichungen

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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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<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 right-hand 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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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>}}
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. Thus the equation can be written as
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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>}}
which we solve by taking roots
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=2</math>,
Zeile 93: Zeile 93:
<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>}}
Complete the square of the left-hand side (use <math>a=-\tfrac{1}{2}</math>)
Complete the square of the left-hand 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>}}
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{{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
and 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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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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Functions
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> and <math>c</math> are constants, and where <math>a\ne0</math>.
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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>}}
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 - Figur - Parabeln y = x² + 2x + 2}}
||{{:2.3 - Figur - Parabeln y = x² + 2x + 2}}
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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, and 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>}}
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{{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
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{{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 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>,
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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>}}
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{{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 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.

Version vom 13:54, 10. Sep. 2008

 

Vorlage:Selected tab Vorlage:Not selected tab

 

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 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 Vorlage:Displayed math 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{,}236 and \displaystyle x=-\sqrt5 \approx -2{,}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 (the 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 that \displaystyle x=1+4=5,
    • \displaystyle x-1 = -\sqrt{16} = -4\, which gives that \displaystyle x=1-4=-3.
  2. Solve the equation \displaystyle \ 2(x+1)^2 -8=0.

    Move the term \displaystyle 8 over to the right-hand side and divide both sides by \displaystyle 2, Vorlage:Displayed math 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, Vorlage:Displayed math and subtract the \displaystyle a^2 from both sides we get

Completing the square: Vorlage:Displayed math

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) Vorlage:Displayed math where the underlined terms are those involved in the completion of the square. Thus the equation can be written as Vorlage:Displayed math which we solve by taking roots
    • \displaystyle x+1 =\sqrt{9} = 3\, and hence \displaystyle x=-1+3=2,
    • \displaystyle x+1 =-\sqrt{9} = -3\, and hence \displaystyle x=-1-3=-4.
  2. Solve the equation \displaystyle \ 2x^2 -2x - \frac{3}{2} = 0.

    Divide both sides by 2 Vorlage:Displayed math Complete the square of the left-hand side (use \displaystyle a=-\tfrac{1}{2}) Vorlage:Displayed math and this gives us the equation Vorlage:Displayed math 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 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:

  • \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 Vorlage:Displayed math has the solutions Vorlage:Displayed 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.

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 left-hand 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

Functions Vorlage:Displayed math are examples of functions of the second degree. In general, a function of the second degree can be written as Vorlage:Displayed math where \displaystyle a, \displaystyle b and \displaystyle c are constants, and where \displaystyle a\ne0.

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

2.3 - Figur - Parablerna y = x² och y = -x²
The figure on the left shows the parabola \displaystyle y=x^2 and figure to the right 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 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

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

    Compared to the parabola \displaystyle y=x^2 all points on the parabola (\displaystyle y=x^2-2) have \displaystyle y-values that are two units smaller, so the parabola has been displaced downwards two units along the \displaystyle y-direction.
2.3 - Figur - Parabeln y = x² - 2
  1. Sketch the parabola \displaystyle \ y=(x-2)^2.

    For the parabola \displaystyle y=(x-2)^2 we need to choose \displaystyle x-values two units larger than for the parabola \displaystyle y=x^2 to get the corresponding \displaystyle y values. So the parabola \displaystyle y=(x-2)^2 has been displaced two units to the right, compared to \displaystyle y=x^2.
2.3 - Figur - Parabeln y = (x - 2)²
  1. Sketch the parabola \displaystyle \ y=2x^2.

    Each point on the parabola \displaystyle y=2x^2 has twice as large a \displaystyle y-value as the corresponding point with the same \displaystyle x-value on parabola \displaystyle y=x^2. Thus parabola \displaystyle y=2x^2 has been increased by a factor \displaystyle 2 in the \displaystyle y-direction as compared to \displaystyle y=x^2.
2.3 - Figur - Parabeln y = 2x²

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

Example 6

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


If one completes the square for the right-hand side Vorlage:Displayed math 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

2.3 - Figur - Parabeln y = x² + 2x + 2

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 Vorlage:Displayed math Complete the square for the left-hand side, Vorlage:Displayed math and this gives the equation Vorlage:Displayed math 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).

2.3 - Figur - Parabeln y = x² - 4x + 3

Example 8

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


We complete the square Vorlage:Displayed math 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.

2.3 - Figur - Parabeln y = x² + 8x + 19


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:

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.


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

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/2.3_Quadratische_Gleichungen“