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secondary 3 | A Maths
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Date Posted: 3 years ago
Views: 446
(√a - √b)²
= (√a)² - 2 √a √b + (√b)²
(recall the property (a - b)² = a² - 2ab + b²)
= a - 2√ab + b
= a + b - 2√ab
So,
√a - √b = √(a + b - 2√ab)
Note that since a > b > 0,
√a > √b > 0
For example, 5 > 4
√5 > √4
So, √a - √b > 0
We have to take the positive square root of (a + b - 2√ab) and not the negative square root.
If you were asked for √b - √a instead, it would equal -√(a + b + √ab)
= (√a)² - 2 √a √b + (√b)²
(recall the property (a - b)² = a² - 2ab + b²)
= a - 2√ab + b
= a + b - 2√ab
So,
√a - √b = √(a + b - 2√ab)
Note that since a > b > 0,
√a > √b > 0
For example, 5 > 4
√5 > √4
So, √a - √b > 0
We have to take the positive square root of (a + b - 2√ab) and not the negative square root.
If you were asked for √b - √a instead, it would equal -√(a + b + √ab)
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Date Posted:
3 years ago
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