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Solutions for quadratic equations by factorisation
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1. The solution for a quadratic equation can be
determined by factoring.
2. Strategy for solving a quadratic equation by
factoring:
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Step 1: Rewrite the equation in standard
form, if necessary.
Step 2: Factorise.
Step 3: Setting each factor equal to 0.
Step 4: Solve each resulting equation in
Step 3.
Step 5: Check the solutions by substituting
back into the original equation.
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Example:
Determine the solutions for each of the followingquadratic equation by factorisation.
a) x281 = 0
b) 5x2 = 45x
c) x2+ 5x24 = 0
d) 16x2 = 8x1
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Solution:
a) x281 = 0
x
2
92
= 0(x9)(x+ 9) = 0
(x9)= 0
x= 9
(x+ 9) = 0
x
= 9
Check:
x281 = 0
LHS:
Whenx= 9,
(9)281 = 81 81
= 0
LHS = RHS
Check:
x281 = 0
LHS:
Whenx= 9,
(9)281 = 81 81
= 0
LHS = RHS
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Solution :
b) 5x2 = 45
5x
2
45 = 05(x29) = 0
5(x232) = 0
5(x3)(x+ 3) = 0
x3 = 0
x= 3
(x+ 3) = 0
x= 3
Check:
5x2= 45
LHS:
Whenx= 3,
5(3)2= 5(9)
= 45
LHS = RHS
Check:
5x2= 45
LHS:
Whenx= 3,
5(3)2= 5(9)
= 45
LHS = RHS
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Solution :
c) x2+ 5x24 = 0
24
1 24
2 12
3 8
x2+ 5x24 = 0
(x3)(x+ 8) = 0
x3 = 0
x= 3
x+ 8 = 0
x= 8
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16x28x+ 1 = 0
(16x4)(16x4) = 0
(4x1)(4x1) = 0
4x1 = 0
4x= 1
x=
Solution :
d) 16x2 = 8x1
16
1 16
2 8
3
4 4
1
4
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Exercise:
1. Solve the equation 3x2= 2(x1) + 7.
15
1 15
2
3 5
3x2= 2(x1) + 7
3x2= 2x2 + 7
3x22x+ 2 7 = 0
3x22x5 = 0
(3x
+ 3)(3x
5) = 0(x+ 1)(3x5) = 0
Solution:x+ 1 = 0
x= 1
3x5 = 0
3x= 5
x
=
5
3
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Exercise:
2. Solve the quadratic equation2m
2
25m
=m +1
4
1 4
2m 1 = 0
2m = 1
m =
m + 2 = 0
m = 2
Solution:
2m2
+ 5m = 2(m + 1)2m2+ 5m = 2m + 2
2m2+ 5m 2m 2 = 0
2m2+ 3m 2 = 0
(2m 1)(2m + 4) = 0
(2m
1)(m + 2) = 0
12
2m2
2
5m=
m +1
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Exercise:
3. Solve the quadratic equation2
2k - 5=3k
3
10
1 10
2k + 1 = 0
2k = 1k =
k 5 = 0
k = 5
Solution:2
2k - 5=3k
3
2k2
5 = 9k2k29k 5 = 0
(2k + 1)(2k 10) = 0
(2k + 1)(k 5)= 0
1-2
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Exercise:
4. Solve the quadratic equation 62
)1(3
x
xx
36
1 36
2 18
3 12
4 9
3x+ 4 = 0
3x= 4
x=
x3 = 0
x= 3
Solution:
62
)1(3
x
xx
3x
2
3x
= 2(x
+ 6)3x23x= 2x+ 12
3x23x2x12 = 0
3x25x12 = 0
(3x+ 4)(3x9) = 0
(3x
+ 4)(x
3) = 0
4-3
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Exercise:
5. Solve the quadratic equation 4x215 = 17x
60
1 60
2 30
3 20
4x3 = 0
4x= 3
x=
x+ 5 = 0
x= 5
Solution:
4x215 = 17x
4x2+ 17x15 = 0
(4x3)(4x+ 20) = 0
(4x3)(x+ 5) = 0
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Summary
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