2.3 Quadratische Gleichungen

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Inhalt:

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

Lernziele:

Nach diesem Abschnitt sollten Sie folgendes können:

  • 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.

Quadratische Gleichungen

Eine quadratische Gleichung kann wie

\displaystyle x^2+px+q=0

geschrieben werden, wo \displaystyle x unbekannt ist, und \displaystyle p und \displaystyle q Konstanten sind.

Einfache quadratische Gleichungen kann man lösen, indem man einfach Wurzeln berechnet.

Die Gleichung \displaystyle x^2=a wo \displaystyle a \ge 0, hat zwei Lösungen (Wurzeln), nämlich \displaystyle x=\sqrt{a} und \displaystyle x=-\sqrt{a}.

Beispiel 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).

Beispiel 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,
    \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

Beispiel 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. Thus the equation can be written as

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

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

    Complete the square of the left-hand 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

    and 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 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

\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.

Beispiel 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

\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 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.

[Image]

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.

Beispiel 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.

[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 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.

[Image]

  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.

[Image]

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

Beispiel 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, 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

[Image]

Beispiel 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).

[Image]

Beispiel 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.

[Image]


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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


Nützliche Websites

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