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secondary 3 | A Maths
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So sorry J keep asking and relying on you
Date Posted: 3 years ago
Views: 167
J how about (ii)
Remember - the sunlight will shine on you one day and give you the knowledge you will receive in the two years of the A Maths syllabus.
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done
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I did this but I can't continue further on
Date Posted:
3 years ago
Correct. Just need to continue.
(32√3 + 8(3) + 196 + 49√3) / (16 - 3)
= (81√3 + 220) / 13
= 81/13 √3 + 220/13
b = 81/13 (or 6 3/13)
a = 220/13 (or 16 12/13)
(32√3 + 8(3) + 196 + 49√3) / (16 - 3)
= (81√3 + 220) / 13
= 81/13 √3 + 220/13
b = 81/13 (or 6 3/13)
a = 220/13 (or 16 12/13)
If you use calculator to check the answer you'll realise both give you the same result
i.e key in (8√3 + 49) ÷ (4 - √3) and compare this with 81/13 √3 + 220/13.
It will be the same.
OR
in one step, key in (8√3 + 49) / (4 - √3) - (81/13 √3 + 220/13)
Since both are supposed to be the same, the result should be 0.
i.e key in (8√3 + 49) ÷ (4 - √3) and compare this with 81/13 √3 + 220/13.
It will be the same.
OR
in one step, key in (8√3 + 49) / (4 - √3) - (81/13 √3 + 220/13)
Since both are supposed to be the same, the result should be 0.
done
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Where did I did wrongly,am not suppose to make it to 2 equations?
Date Posted:
3 years ago
Again, you didn't simplify the equations.
8c√3 + c² = 8√3 + 49 - 48
8c√3 + c² = 8√3 + 1
Comparing coefficients of √3,
8c = 8
c = 1
Comparing coefficients of c²,
c² = 1
c = 1 or = -1(rejected as that would make 8c = -8, which is negative andwould not equal the 8 on the RHS)
8c√3 + c² = 8√3 + 1
Comparing coefficients of √3,
8c = 8
c = 1
Comparing coefficients of c²,
c² = 1
c = 1 or = -1(rejected as that would make 8c = -8, which is negative andwould not equal the 8 on the RHS)
Alternatively,
8c√3 + c² = 8√3 + 1
c² + 8√3 c - (8√3 + 1) = 0
(c - 1)(c + 8√3 + 1) = 0
c - 1 = 0 or c + 8√3 + 1 = 0
c = 1 or c = -8√3 - 1 (rejected , as this would mean the side of the square = 4√3 - 8√3 - 1
= -4√3 - 1, which is negative and we know that length cannot be negative)
So, c = 1
8c√3 + c² = 8√3 + 1
c² + 8√3 c - (8√3 + 1) = 0
(c - 1)(c + 8√3 + 1) = 0
c - 1 = 0 or c + 8√3 + 1 = 0
c = 1 or c = -8√3 - 1 (rejected , as this would mean the side of the square = 4√3 - 8√3 - 1
= -4√3 - 1, which is negative and we know that length cannot be negative)
So, c = 1
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