2.3 Quadratische Gleichungen
Aus Online Mathematik Brückenkurs 1
Content:
- Completing the square method
- Quadratic equations
- Faktorising
- 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:Fristående formel 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{,}236 and \displaystyle x=-\sqrt5 \approx -2{,}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 (the 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 that \displaystyle x=1+4=5,
- \displaystyle x-1 = -\sqrt{16} = -4\, which gives that \displaystyle x=1-4=-3.
- 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:Fristående formel 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:Fristående formel and subtract the \displaystyle a^2 from both sides we get
Completing the square: Vorlage:Fristående formel
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) Vorlage:Fristående formel where the underlined terms are those involved in the completion of the square. Thus the equation can be written as Vorlage:Fristående formel 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.
- Solve the equation \displaystyle \ 2x^2 -2x - \frac{3}{2} = 0.
Divide both sides by 2 Vorlage:Fristående formel Complete the square of the left-hand side (use \displaystyle a=-\tfrac{1}{2}) Vorlage:Fristående formel and this gives us the equation Vorlage:Fristående formel Taking roots gives- \displaystyle x-\tfrac{1}{2} =\sqrt{1} = 1, \quad dvs. \displaystyle \quad x=\tfrac{1}{2}+1=\tfrac{3}{2},
- \displaystyle x-\tfrac{1}{2}= -\sqrt{1} = -1, \quad dvs. \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 for LH and RH respectively:
- \displaystyle x = 2 gives att \displaystyle \mbox{LH } = 2^2 +2\cdot 2 - 8 = 4+4-8 = 0 = \mbox{RH}.
- \displaystyle x = -4 gives att \displaystyle \mbox{LH } = (-4)^2 + 2\cdot(-4) -8 = 16-8-8 = 0 = \mbox{RH}.
In both cases we arrive at LH = RH. The equation is satisfied in both cases.
Using the completing the square method it is possible to show that the general quadratic equation Vorlage:Fristående formel has the solutions Vorlage:Fristående formel 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 factorise 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 Vorlage:Fristående formel are examples of functions of the second degree. In general, a function of the second degree can be written as Vorlage:Fristående formel where \displaystyle a, \displaystyle b and \displaystyle c are constants, and where \displaystyle a\ne0.
The graph for a functions 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.
As the expression \displaystyle x^2 is minimum 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
| 2.3 - Figur - Parabeln y = x² - 2 |
| 2.3 - Figur - Parabeln y = (x - 2)² |
| 2.3 - Figur - Parabeln y = 2x² |
All sorts of parabolas can be handled by the completing the square method.
Exempel 6
| Sketch the parabola \displaystyle \ y=x^2+2x+2.
| 2.3 - Figur - Parabeln y = x² + 2x + 2 |
Example 7
Determine where the parabola \displaystyle \,y=x^2-4x+3\, cuts \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:Fristående formel
Complete the square for the left-hand side,
Vorlage:Fristående formel
and this gives the equation
Vorlage:Fristående formel
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
Vorlage:Fristående formel
and then we see that the term 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 xx=-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:
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
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
