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secondary 3 | A Maths
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Date Posted: 3 years ago
Views: 320
A more condensed answer with fewer steps but ensures full credit :
①
a³ = ( ³√(2+√5) + ³√(2-√5) )³
②
a³ = (³√(2+√5))³
+ 3(³√(2+√5))² (³√(2-√5))
+ 3(³√(2+√5)) (³√(2-√5))²
+ (³√(2-√5))³
③
a³ = 2 + √5
+3(³√(2-√5)) (³√(2+√5)) (³√(2+√5) + ³√(2-√5))
+ 2 - √5
④ a³ = 4 + 3a ³√[(2-√5)(2+√5)]
⑤ a³ = 4 + 3a ³√(2²-(√5)²)
⑥ a³ = 4 + 3a ³√-1
⑦ a³ = 4 + 3a(-1)
⑧ a³ = 4 - 3a (shown)
Next part :
a³ + 3a - 4 = 0
When a = 1,
1³ + 3(1) - 4 = 1 + 3 - 4 = 0
By the factor theorem, (a - 1) is a factor of a³ + 3a - 4
Factorise,
(a - 1)(a² + a + 4) = 0
a - 1 = 0 → a = 1
Or
a² + a + 4 = 0
Discriminant = 1² - 4(1)(4) = -15 < 0
There are no real roots for a² + a + 4 = 0
∴ a = 1
①
a³ = ( ³√(2+√5) + ³√(2-√5) )³
②
a³ = (³√(2+√5))³
+ 3(³√(2+√5))² (³√(2-√5))
+ 3(³√(2+√5)) (³√(2-√5))²
+ (³√(2-√5))³
③
a³ = 2 + √5
+3(³√(2-√5)) (³√(2+√5)) (³√(2+√5) + ³√(2-√5))
+ 2 - √5
④ a³ = 4 + 3a ³√[(2-√5)(2+√5)]
⑤ a³ = 4 + 3a ³√(2²-(√5)²)
⑥ a³ = 4 + 3a ³√-1
⑦ a³ = 4 + 3a(-1)
⑧ a³ = 4 - 3a (shown)
Next part :
a³ + 3a - 4 = 0
When a = 1,
1³ + 3(1) - 4 = 1 + 3 - 4 = 0
By the factor theorem, (a - 1) is a factor of a³ + 3a - 4
Factorise,
(a - 1)(a² + a + 4) = 0
a - 1 = 0 → a = 1
Or
a² + a + 4 = 0
Discriminant = 1² - 4(1)(4) = -15 < 0
There are no real roots for a² + a + 4 = 0
∴ a = 1
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Date Posted:
3 years ago
The formula (a + b)³ = a³ + 3a²b + 3ab² + b³ can be used straight away since it's in the A Math syllabus. No need to incur extra steps expanding to (a + b)²(a + b) and expand out
Furthermore, substitution of a = 1 to arrive at a³ + 3a - 4 = 0, (then quoting the factor theorem) has to be shown for full credit.
Writing 'By inspection, ... ... ' will not suffice since it is not immediately obvious that f(-1) will lead to 0 and evaluation is needed
This is unlike the following :
4x³ + 8x² + x - 3 = (x + 1)(Ax² + Bx + C)
where it is intuitive that the only term in x³ is x(Ax²) so by inspection/observation,
A = 4
Likewise, C = -3
Writing 'By inspection, ... ... ' will not suffice since it is not immediately obvious that f(-1) will lead to 0 and evaluation is needed
This is unlike the following :
4x³ + 8x² + x - 3 = (x + 1)(Ax² + Bx + C)
where it is intuitive that the only term in x³ is x(Ax²) so by inspection/observation,
A = 4
Likewise, C = -3
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