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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

34

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Summary

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