Understanding Engineering Mathematics

are factors and we have

(x−α)(x−β)≡x^2 +ax+b

Expanding the left-hand side gives

x^2 −(α+β)x+αβ≡x^2 +ax+b

So
b=product of roots=αβ
a=−sum of roots=−(α+β)

Solution to review question 2.1.11

A.(i) You may have jumped straight in to the formula here. But, in fact,

all we have to do is take out anxto give

x^2 − 3 x=x(x− 3 )= 0

The only possible solutions are then

x=0andx=3 (don’t forget the zero solution!)

i.e.

x=0or3

(ii) A routine factorisation gives

x^2 − 5 x+ 6 =(x− 3 )(x− 2 )= 0

so

x=2or3

(iii) This sort of problem, with a coefficient ofx^2 that is not one

sometimes gives problems when it comes to using factorisation.

We have to try factors of both the 2 and the−2. Some trial and

error leads to

2 x^2 + 3 x− 2 ≡( 2 x− 1 )(x+ 2 )= 0

So we get solutionsx=^12 ,− 2

If you do get stuck on the factorisation then use the formula –but

be careful in handling the coefficient ofx^2. The solution of

ax^2 +bx+c= 0

is

x=

−b±

√

b^2 − 4 ac

2 a