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secondary 3 | A Maths
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please help me with part 2 :-)
Date Posted: 5 years ago
Views: 298
a + b√3 = (5 + 2√3)²/(2 + √3)
a + b√3 = (25 + 20√3 + 12)/(2 +√3)
a + b√3 = (37 + 20√3)/(2 + √3) x (2 - √3)/(2 - √3)
a + b√3 = (37 + 20√3)(2 - √3)/(2² - (√3)²)
a + b√3 = (74 - 37√3 + 40√3 - 60)/(1)
a + b√3 = 14 + 3√3
Comparing coefficients,
a = 14
b = 3
Idea here is to bring over the denominator 5 + 2√3 to the other side.
Then rationalise the denominator 2 +√3
Some steps were skipped/simplified
a + b√3 = (25 + 20√3 + 12)/(2 +√3)
a + b√3 = (37 + 20√3)/(2 + √3) x (2 - √3)/(2 - √3)
a + b√3 = (37 + 20√3)(2 - √3)/(2² - (√3)²)
a + b√3 = (74 - 37√3 + 40√3 - 60)/(1)
a + b√3 = 14 + 3√3
Comparing coefficients,
a = 14
b = 3
Idea here is to bring over the denominator 5 + 2√3 to the other side.
Then rationalise the denominator 2 +√3
Some steps were skipped/simplified
See 1 Answer
done
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clear
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Hi Rachel! Here are my workings for the two questions
Date Posted:
5 years ago
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