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secondary 4 | E Maths
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Date Posted: 5 years ago
Views: 273
2x² - 6x - k = 0
Sum of roots
= -(-6)/2
= 3
Since roots differ by 5,
First root = second root + 5
Sum = first root + second root
= Second root + 5 + second root
= 2 x second root + 5
2 x second root + 5 = 3
So 2 x second root = 3 - 5 = -2
Second root = -2 ÷ 2 = -1
First root = -1 + 5 = 4
(You can also use first root = second root - 5. The roots will just swap places.
i.e first root will then be -1 and second root will be 4 )
Product of roots
= -k/2= 4 x (-1) = -4
k/2 = 4
k = 2 x 4 =
To check (if you wanna be sure. Or you can also use this as the alternate working for finding k)
x = 4 or x = -1
x - 4 = 0 or x + 1 = 0
So since they are the two roots,
(x - 4)(x + 1) = 0
x² - 3x - 4 = 0
2x² - 6x - 8 = 0
k = 8
Sum of roots
= -(-6)/2
= 3
Since roots differ by 5,
First root = second root + 5
Sum = first root + second root
= Second root + 5 + second root
= 2 x second root + 5
2 x second root + 5 = 3
So 2 x second root = 3 - 5 = -2
Second root = -2 ÷ 2 = -1
First root = -1 + 5 = 4
(You can also use first root = second root - 5. The roots will just swap places.
i.e first root will then be -1 and second root will be 4 )
Product of roots
= -k/2= 4 x (-1) = -4
k/2 = 4
k = 2 x 4 =
To check (if you wanna be sure. Or you can also use this as the alternate working for finding k)
x = 4 or x = -1
x - 4 = 0 or x + 1 = 0
So since they are the two roots,
(x - 4)(x + 1) = 0
x² - 3x - 4 = 0
2x² - 6x - 8 = 0
k = 8
See 1 Answer
done
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Alternatively, with no knowledge of a maths, you can attempt this by using general equation of quadratic
Date Posted:
5 years ago
Thank you
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