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junior college 2 | H3 Maths
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pls help!
Date Posted: 4 years ago
Views: 605
Rationalise both the numerator and denominator.
(√(8 + x) - 3 ) / (√(3 + x) - 2)
= (√(8 + x) - 3 ) / (√(3 + x) - 2) x (√(3 + x) + 2)/(√(3 + x) + 2) x (√(8 + x) + 3)/(√(8 + x) + 3)
= (8 + x - 3²)(√(3 + x) + 2)/ [(3 + x - 2²)(√(8 + x) + 3) ]
(Using the property ((a + b)(a - b) = a² - b²)
= (x - 1)(√(3 + x) + 2) / [(x - 1)(√(8 + x) + 3)]
= (√(3 + x) + 2)/ (√(8 + x) + 3)
Last edit 2.32am
(√(8 + x) - 3 ) / (√(3 + x) - 2)
= (√(8 + x) - 3 ) / (√(3 + x) - 2) x (√(3 + x) + 2)/(√(3 + x) + 2) x (√(8 + x) + 3)/(√(8 + x) + 3)
= (8 + x - 3²)(√(3 + x) + 2)/ [(3 + x - 2²)(√(8 + x) + 3) ]
(Using the property ((a + b)(a - b) = a² - b²)
= (x - 1)(√(3 + x) + 2) / [(x - 1)(√(8 + x) + 3)]
= (√(3 + x) + 2)/ (√(8 + x) + 3)
Last edit 2.32am
Limit
= (√(3 + 1) + 2)/(√(8 + 1) + 3)
= (√4 + 2)/(√9 + 3)
= (2 + 2)/(3 + 3)
= 4/6
= ⅔
Last edit 2.22am
= (√(3 + 1) + 2)/(√(8 + 1) + 3)
= (√4 + 2)/(√9 + 3)
= (2 + 2)/(3 + 3)
= 4/6
= ⅔
Last edit 2.22am
oic thanks!
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Here's a similar question posted by another student (from NTU I think) a few weeks back.
Date Posted:
4 years ago
NUS*
The answer is 16/27.
But it's a little trickier as after rationalising both numerator and denominator, you'll get a cubic polynomial on both the numerator and denominator, which you'll need to factorise, and then cancel out common factors.
But it's a little trickier as after rationalising both numerator and denominator, you'll get a cubic polynomial on both the numerator and denominator, which you'll need to factorise, and then cancel out common factors.
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