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junior college 1 | H2 Maths
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Good evening! How do i solve this? Just part i will do
Date Posted: 4 years ago
Views: 271
(x² - 4x)/(x² - 9)
= (x² - 9 - 4x + 9)/(x² - 9)
= 1 + (9 - 4x)/(x² - 9)
= 1 + (9 - 4x)/[(x + 3)(x - 3)]
Let (9 - 4x)/[(x + 3)(x - 3)] = Q/(x - 3) + R/(x + 3)
9 - 4x = Q(x + 3) + R(x - 3)
Sub x = -3,
9 - 4(-3) = R(-3 - 3)
21 = -6R
R = -21/6 = -7/2
Sub x = 3,
9 - 4(3) = Q(3 + 3)
-3 = 6Q
Q = -3/6 = -½
So,
(x² - 4x)/(x² - 9) = 1 + (-½/(x - 3)) + (-7/2 / (x + 3) )
= (x² - 9 - 4x + 9)/(x² - 9)
= 1 + (9 - 4x)/(x² - 9)
= 1 + (9 - 4x)/[(x + 3)(x - 3)]
Let (9 - 4x)/[(x + 3)(x - 3)] = Q/(x - 3) + R/(x + 3)
9 - 4x = Q(x + 3) + R(x - 3)
Sub x = -3,
9 - 4(-3) = R(-3 - 3)
21 = -6R
R = -21/6 = -7/2
Sub x = 3,
9 - 4(3) = Q(3 + 3)
-3 = 6Q
Q = -3/6 = -½
So,
(x² - 4x)/(x² - 9) = 1 + (-½/(x - 3)) + (-7/2 / (x + 3) )
Thank you!
Welcome. The concept used here is partial fractions.
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done
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clear
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When you see the power of x in the denominator is the same as the power of x in the numerator,the first thing you should do is to simplify it to proper fraction. Then using the hint they provided in the qn. Use partial fractions and further simplify b4 going on to the last part
Date Posted:
4 years ago
Slight error in the final answer. The negative sign from '1 - ' applies to both fractions.
It should be 1 - 1/(2(x - 3)) - 7/(2(x + 3))
= 1 + (-½/(x - 3)) + (-7/2 / (x + 3) ), which is in the form the question specified
= 1 + (-½/(x - 3)) + (-7/2 / (x + 3) ), which is in the form the question specified
Thank you!
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