year 8 maths

2F
Solving simple quadratic equations
89
PROBLEM-SOLVING AND REASONING
7, 8, 10 7, 9(½), 11, 12 9(½), 10, 12, 13
7 The area of a square is 25 m 2 . Find its perimeter.
8 A square mirror has an area of 1 m 2 . Find its perimeter.
9 By first dividing both sides by the coefficient of x 2 , solve these simple quadratic equations.
a 2x 2 = 8 b 3x 2 = 3
c 5x 2 = 45 d −3x 2 = −12
e −2x 2 = −50 f 7x 2 = 0
g −6x 2 = −216
h −10x 2 = −1000
10 Explain why:
a x 2 = 0 has only one solution b x 2 = c has no solutions when c < 0 .
11 Solve the equations with the given conditions.
a x 2 = 16 when x > 0 b x 2 = 25 when x < 0
c x 2 = 49 when x < 0 d x 2 = 196 when x > 0
12 The exact value solutions to x 2 = 5 , for example, are written as x = √5 or −√5 . Alternatively,
we can write x = ±√5 .
Write the exact value solutions to these equations.
a x 2 = 11 b x 2 = 17
c x 2 = 33 d x 2 = 156
13 The triad (a, b, c) = (3, 4, 5) satisfies the equation a 2 + b 2 = c 2 because 3 2 + 4 2 = 5 2 .
a Decide if the following triads also satisfy the equation a 2 + b 2 = c 2 .
i (6, 8, 10) ii (5, 12, 13)
iii (1, 2, 3) iv (−3, −4, −5)
v (−2, −3, −6) vi (−8, 15, 17)
b Can you find any triads, for which a, b, c are positive integers, that satisfy a 3 + b 3 = c 3 ?
ENRICHMENT
– –
14
Solving more complex linear equations
14 Compare this linear and quadratic equation solution.
3x − 1 = 11
3x
+1
− 1 = 11
+1 +1
+1
3x = 12
3x 2 = 12
÷3 ÷3 ÷3
÷3
x = 4
x 2 = 4
x = ±2
Now solve these quadratic equations.
a 2x 2 + 1 = 9 b 5x 2 − 2 = 3
c 3x 2 − 4 = 23 d −x 2 + 1 = 0
e −2x 2 + 8 = 0 f 7x 2 − 6 = 169
g 4 − x 2 = 0 h 27 − 3x 2 = 0
i 38 − 2x 2 = −34
Cambridge Maths NSW
Stage 4 Year 8 Second edition
ISBN 978-1-108-46627-1 © Palmer et al. 2018
Cambridge University Press
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