Page 4 - Equations, substitution formulae
P. 4
c. ( + 9)( + 11) = 0 d. ² + 4 = 0
ℎ + 9 = 0 + 11 = 0 Factorise first
= − 9 = −11 ( + 4) = 0
ℎ = 0 + 4 = 0
= 0 = − 4
e. ² + 6 + 9 = 0 e. ² + 2 − 15 = 0
( + 3)( + 3) = 0 ( – 3)( + 5) = 0
ℎ + 3 = 0 ( ) ℎ – 3 = 0 + 5 = 0
= − 3 ( ) = 3 = − 5
Substitution
Examples
a. Given that = 4 and = 3. Determine the value of i. + ii. iii.
² − ³ ²
b. Given that = −3, = −5 and = 4. Determine i. ii. ² − ³ +
Solutions
a = 4 and = 3.
i. + = 4 + 3 ii. ab = 4 × 3 iii. ² − ³ ² = 4² × 3 −
4³ × 3²
= 7 = 12 = (4×4×3) -
(4×4×4×3×3)
= 48 – 576
= - 528
b. = −3, = −5 and = 4.
i. = −3 × −5 × 4 = 60 ii. ² − ³ + = (−3)² − (−5)² +
4
=
(−3)(−3) – (−5)(−5) + 4
= 3
