quadratic_equations.pdf

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FACULTY OF

BUSINESS &

ECONOMICS

QUADRATICEQUATIONS

This publication can be cited as: Carter, D. (2008),

Quadratic Equations, Teaching and Learning Unit,

Faculty of Business and Economics, the University of

Melbourne. http://tlu.fbe.unimelb.edu.au/

Further credits: Beaumont, T. (content changes and

editing), Pesina, J. (design and layout).

Helpsheet

Use this sheet to help you:

Recognize a quadratic expression and how to expand or factorise itSolve a quadratic expression using trial-and-error methods and also byusing the formula

Understand the relationship between the algebraic and graphicalsolution of a quadratic equationRecognize and solve simultaneous quadratic equations


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Helpsheet

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QUADRATIC EQUATIONS

Quadratic expressions

Expanding pairs of brackets

(a + b)(c + d),(wherea,b,canddareunspeciedconstants)= ac + ad + bc + bd.

*One way to help remember this is FOIL First, Outside, Inside, Last.

Many expressions you will see are of the form (x + a)(x + b), multiplying this out givesx2 + (a + b)x + ab

A quadratic expression contains an unknown x raised to the power 2, no higher orlower.

Technique for factorising a quadratic expression

Remember: factorising is the reverse of expanding.e.g Factorise x2 + 12x + 32

Assume the factors are (x + a)(x + b).

From this (x + a)(x + b) = x2 + (a + b)x + ab. From the expression givena + b = 12 and ab = 32. So look for two numbers whose sum is 12 and product is 32.

Bytrialanderrorwendthat4+8=12and4x8=32,sowehavethefactors(x+4)and(x + 8).

Therefore the solution to the factorisation ofx2 + 12x + 32 is (x+4)(x + 8).

Not all expressions can be factorised using this technique.

Quadratic equations

We are asked to solve the equationx2 + 12x + 32 = 0

From the previous example:x2 + 12x + 32= (x+4)(x + 8)

Therefore we need to solve (x +4)(x + 8) = 0.

When ab =0, then eithera = 0 orb = 0,

So eitherx +4=0orx + 8 = 0, which gives x=-4andx = -8 as solutions (also known asroots).

You can check your answers by substituting each one in turn into the original equation.

The general form of a quadratic equation isax2 + bx + c = 0


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Helpsheet

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QUADRATIC EQUATIONS

The formula for solving any quadratic equation

Usingthetrialanderrormethodcanbequitedifcultandtimeconsumingandreliesheavily on guesswork. A formula exists for solving any quadratic equation. This takes usstraight to the roots and will also tell us whether a solution exists.

Given any quadratic equation

ax2 + bx + c = 0 (where a, b and c are given constants) the solution (roots) are given bythe formula

x = - b (b2- 4ac)2a

e.g x2 + 12x + 32 = 0 from this we have a = 1, b = 12 and c = 32,

substituting these values into the formula gives

x=-12(1224x1x32)2 x 1

=-12(144128)2

=-12162

x =-12+4or x=-12-42 2

x=-4orx=-8

(as found previously using trial and error method)In this example we found two roots, from this we can deduce that if b2-4ac > 0, therewill be two solutions

Not all quadratics can be factorised. If b2-4acisanegativeresult,thenwehavetonda square root of a negative number, which we have seen earlier it has no square root.Therefore if b2-4ac is negative there are no real roots to the equation. i.e. If b2-4ac < 0,there is no solution.

Some quadratics have only one root (solution). These are called perfect squares.e.g. x2 + 10x + 25 = 0.Usingtrialanderrorortheformulayouwillnd(x + 5) (x + 5) = 0 or (x + 5)2 = 0=> x = -5A perfect square occurs when b2-4ac = 0. i.e. there will be one solution.

Quadratic Functions

y = x2 + 12x + 32 is an example of a quadratic function.

The general form of a quadratic function isy = ax2+ bx + c. (a, b, c are parameters)

The graph of a quadratic function is a curve called a parabola. It is a U-shape arisingfrom the fact thatx2 is positive whenx is either positive or negative. If the parametera is positive then the U-shape has its two arm pointing upwards. If the parameter a isnegative, then the U-shape has its two arms pointing downwards. The absolute value ofa determines how steeply the curve turns up (or down).


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