Understanding Engineering Mathematics

side is always the same as the total degree of the original denominator. In the above the
cover-up rule may be used forA,B,Cin the first result, forAin the second and forAand
Din the third. There are many examples of the more complicated types in the reinforcement
exercises, but the simplest case of linear factors should cover most of your needs.

Solution to review question 2.1.10

Write

x+ 1

(x− 1 )(x+ 3 )

≡

A

x− 1

+

B

x+ 3

≡

A(x+ 3 )+B(x− 1 )

(x− 1 )(x+ 3 )

so

A(x+ 3 )+B(x− 1 )≡x+ 1

You can now determineAandBfrom this identity by substitutingx= 1

to giveA=^12 andx=−3togiveB=^12. Or, equate coefficients and

solve the equations

A+B= 1

3 A−B= 1

Check these results by using the ‘cover-up rule’. We finally get

x+ 1

(x− 1 )(x+ 3 )

≡

1

2 (x− 1 )

+

1

2 (x+ 3 )

2.2.11 Properties of quadratic expressions and equations

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39 78 ➤

So far, we have dealt mainly with algebraic expressions and rearranging them – now we
turn to solving simple algebraic equations. We will only be considering solution byexact
rather thannumericalmeans here.
The first point to emphasise is that an equation is a precise statement involving an
equality, and any solution must reflect that precision, or else it is only an approximate
solution. Consider for example the simple equation

3 x− 2 = 0

The precise solution is

x=

2

3

If you now use your calculator to give you the approximationx= 0 .666666667, then
this isnota solution of the equation. It is simply an approximation to it and you should
write your answer asx
0 .666666667. You may feel that this is nit-picking, with no real
practical importance, since we are always using approximations in engineering and science.
However, small errors can often assume serious significance in engineering systems, and