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secondary 4 | A Maths
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Pls help for 20 (iii).
Thank you very much.
= 34 - 2(12√2)
= 16 - 2(4)(3√2) + 18
= 4² - 2(4)(3√2) + (3√2)²
= (4 - 3√2)²
(Recall (a - b)² = a² - 2ab + b²)
So one of the square roots is 4 - 3√2
The other is the negative of it
-(4 - 3√2) = 3√2 - 4
Alternatively,
Let 34 - 24√2 = (a - b√2)² = a² - 2ab√2 + 2b²
comparing coefficients,
a² + 2b² = 34 ①
2ab = 24 ②
ab = 12
a = 12/b ③
Sub ③ into ①,
(12/b)² + 2b² = 34
144/b² + 2b² = 34
144 + 2b⁴ = 34b²
2b⁴ - 34b² + 144 = 0
b⁴ - 17b² + 72 = 0
(b² - 8)(b² - 9) = 0
b² = 8 or b² = 9
b = ±√8 or b = ±3
b = ±2√2
a = ±12/(2√2) or a = ±12/3
a = ± 6/√2 or a = ± 4
a = ±6√2/2
a = ± 3√2
(Notice if b = ±2√2, then b√2 = ±4.
So in either case you will get ±(4 - 3√2) as your roots
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