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secondary 3 | A Maths
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secondary 3 chevron_right A Maths chevron_right Singapore

help pls, thx!

Date Posted: 4 years ago
Views: 160
J
J
4 years ago
properties to use :

① a³ - b³ = (a – b)(a² + ab + b²)

②a² - b² = (a - b)(a + b)

③(a + b)² = a² + 2ab + b²

So x^6 - 64

= (x²)³ - 4³

= (x² - 4)((x²)² + x²(4) + 4²) ( property ①)

= (x + 2)(x - 2)( (x²)² + 4x² + 4x² + 4² - 4x²) (property②)

= (x + 2)(x - 2)( (x²)² + 2(x²)(4) + 4² - 4x²)

(Completing the square)


= (x + 2)(x - 2)( (x² + 4)² - (2x)²)
(applying property ③)

= (x + 2)(x - 2)(x² + 4 - 2x)(x² + 4 + 2x)
(Property ② again)


ii)

x^6 - 64 = (x² + 4)² - 4x²

So (x + 2)(x - 2)(x² + 4 - 2x)(x² + 4 + 2x) = (x² + 4)² - 4x²


(x + 2)(x - 2)(x² + 4 - 2x)(x² + 4 + 2x) = (x² + 4 - 2x)(x² + 4 + 2x)

(from i)

(x + 2)(x - 2)(x² + 4 - 2x)(x² + 4 + 2x) - (x² + 4 - 2x)(x² + 4 + 2x) = 0


( (x + 2)(x - 2) - 1)(x² + 4 - 2x)(x² + 4 + 2x) = 0

((x + 2)(x - 2) - 1) = 0 or x² + 4 - 2x = 0 or x² + 4 + 2x = 0

x² - 4 - 1 = 0 or x² + 4 - 2x = 0 or x² + 4 + 2x = 0

x² - 5 = 0 or x² - 2x + 1 = -3 or x² + 2x + 1 = -3

x² = 5 or (x - 1)² = -3 or (x + 1)² = -3

(Second and third case are rejected as (x - 1)² and (x + 1)² ≥ 0 for all real x. The square of a real value is either 0 or positive)

x = √5 or x = -√5
help
Help
4 years ago
tysm!
J
J
4 years ago
Np

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