Cambridge International Mathematics

(Tina Sui) #1
Try as much as we like, we will not be able to solve quadratic
equations such asx^2 +4x¡7=0using the factorisation methods
already practised. This is because the solutions are not rationals.

Not all
quadratics have
simple factors.

Consequently, thequadratic formulahas been developed:

If ax^2 +bx+c=0 where a 6 =0, then x=

¡b§

p
b^2 ¡ 4 ac
2 a

.

Proof: If ax^2 +bx+c=0

then x^2 +

b
a

x+

c
a

=0 fdividing each term bya,asa 6 =0g

) x^2 +

b
a

x =¡

c
a

) x^2 +
b
a

x+

μ
b
2 a

¶ 2

c
a

+

μ
b
2 a

¶ 2
fcompleting the square on LHSg

)

μ
x+

b
2 a

¶ 2

c
a

μ
4 a
4 a


+

b^2
4 a^2

)

μ
x+
b
2 a

¶ 2
=
b^2 ¡ 4 ac
4 a^2

) x+

b
2 a


r
b^2 ¡ 4 ac
4 a^2

) x=¡
b
2 a

§

p
b^2 ¡ 4 ac
2 a

) x=
¡b§

p
b^2 ¡ 4 ac
2 a
To demonstrate the validity of this formula, consider the equation x^2 ¡ 3 x+2=0.

By factorisation: x^2 ¡ 3 x+2=0
) (x¡1)(x¡2) = 0
) x=1or 2

By formula: a=1,b=¡ 3 ,c=2

) x=

¡(¡3)§

p
(¡3)^2 ¡4(1)(2)
2
) x=

3 §

p
9 ¡ 8
2
) x=

3 § 1

2

) x=2or 1

We can see that factorisation is quicker if the quadratic can indeed be factorised, but the quadratic formula
provides an alternative for when it cannot be factorised.

C THE QUADRATIC FORMULA [2.10]


Quadratic equations and functions (Chapter 21) 427

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