Understanding Engineering Mathematics

so in this case we get

x=

− 3 ±

√

32 − 4 × 2 ×(− 2 )

2 × 2

=

− 3 ±

√

25

4

=

− 3 ± 5

4

=^12 or − 2 , as above

B. To complete the square we proceed as follows (see text above):

• if necessary factor out the coefficient ofx^2


• take half the coefficient ofx


• square it


• add and subtract the result to the quadratic expression


• use the result(x+a)^2 =x^2 + 2 ax+a^2 with the added bit

In other words ‘add and subtract(half of the coefficient ofx)^2

′

’. This

does not change the value of the expression, because we are simply

adding zero, but it sets up the expression for us to use the result

(x+a)^2 =x^2 + 2 ax+a^2. Thus we have

x^2 +x+ 1 ≡x^2 +x+ 1 +

( 1

2

) 2

−

( 1

2

) 2

≡x^2 +x+

( 1

2

) 2

+^34

=

(

x+^12

) 2

+^34

This form of the expression makes it clear that its minimum value must

be^34 , which must occur whenx+^12 =0, i.e.x=−^12.

C. Ifα,βare the roots of the quadraticx^2 + 2 x+3 then their sum is

the negative of the coefficient ofx, while their product is the constant

term (including sign). So, we have

α+β=− 2

αβ= 3

2.2.12 Powers and indices for algebraic expressions

➤

40 78 ➤

We introduced powers and indices in Section 1.2.7, mainly for numbers. We will extend them
here to algebraic symbols. The rules, reproduced here, are exactly the same for algebraic
functions:

aman=am+n

am

an

=am−n

(am)n=amn