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Revision:Quadratic Equations

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A quadratic equation is an equation where the highest power of x is x^2. There are various methods of solving quadratic equations, as shown below.


NOTE: If x^2 = 36, then x = +6 or -6 (since squaring either of these numbers will give 36).

However, \sqrt{36} = +6 only - negative numbers do not have 'nice' square roots so we don't deal with them at this level.


Contents

Completing the Square

Our first method for solving quadratic equations.

9 and 25 can be written as 3² and 5² whereas 7 and 11 cannot be written as the square of another exact number. 9 and 25 are called perfect squares. Another example is:

\displaystyle \frac{9}{4} = (\frac{3}{2})^2.

In a similar way:

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


To make x^2 + 6x into a perfect square, we add \frac{6^2}{4} = 9. The resulting expression:

\displaystyle x^2 + 6x + 9 = (x + 3)^2

and so is a perfect square. This is known as completing the square. To complete the square in this way, we take the number before the x, square it, and divide it by 4. This technique can be used to solve quadratic equations, as demonstrated in the following example.


Example

Solve x^2 - 6x + 2 = 0 by completing the square

\displaystyle x^2 - 6x = -2

[To complete the square on the LHS (left hand side), we must add \frac{6^2}{4} = 9. We must, of course, do this to the RHS also]. So:

\displaystyle x^2 - 6x + 9 = 7

 (x - 3)^2 = 7

[Now take the square root of each side]

\displaystyle x - 3 = \pm2.646 (the square root of 7 is +2.646 or -2.646)

\displaystyle x = 5.646 or \displaystyle 0.354


Completing the square can also be used to find the maximum or minimum point on a graph.

Example

Find the minimum of the graph y = 3x^2 - 6x - 3.

In this case, the x^2 has a '3' in front of it, so we start by taking the three out:

\displaystyle  y = 3(x^2 - 2x -1).

[This is the same since multiplying it out gives 3x^2 - 6x - 3]

Now complete the square for the bit in the bracket:

\displaystyle y = 3[(x - 1)^2 - 2]

Multiply out the big bracket:

\displaystyle y = 3(x - 1)^2 - 6

We are trying to find the minimum value that this graph can be. (x - 1)^2 must be zero or positive, since squaring a number always gives a positive answer. So the minimum value will occur when (x - 1)^2 = 0, which is when x = 1. When x = 1, y = -6. So the minimum point is at (1, -6).

Some people don't like the method of completing the square to solve equations and an alternative is to use the quadratic formula. This is actually derived by completing the square.


The Quadratic Formula

The quadratic formula is


\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}


It gives you the solutions to the a general quadratic equations:

\displaystyle ax^2 + bx + c = 0.


Example

Solve: 3x^2 + 5x - 8 = 0

\displaystyle x = \frac{-5 \pm \sqrt{5^2 - (4\times 3\times (-8))}}{2\times 3}

\displaystyle = \frac{-5 \pm \sqrt{25 + 96}}{6}

\displaystyle = \frac{-5 \pm \sqrt{121}}{6}

\displaystyle = \frac{-5 + 11}{6} or \displaystyle \frac{-5 - 11}{6}

so: \displaystyle x = 1 or \displaystyle -2.33


Factorising

Sometimes, quadratic equations can be solved by factorising. In this case, factorising is probably the easiest way to solve the equation.


Example

Solve: x^2 + 2x - 8 = 0

\displaystyle (x - 2)(x + 4) = 0


either \displaystyle x - 2 = 0 or \displaystyle x + 4 = 0

so\displaystyle x = 2 or \displaystyle x = - 4


If you do not understand the third line, remember that for (x - 2)(x + 4) to equal zero, then one of the two brackets must be zero.

If a question asks for you to round your answers to 2 decimal places (2dp) then it is probably not a factorising question and instead you probably have to use the quadratic formula.

Comments

Either more needs to be made on the factorising method in this article or a link made to the factorising article is added in and that article is developed significantly.

A diagram is needed for the quadratic formula section. Also, a graph could be useful for the 'completing the square to find max and min points' section.

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