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junior college 1 | H2 Maths
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Candice
Candice

junior college 1 chevron_right H2 Maths chevron_right Singapore

Good evening! How do i solve this? Just part i will do

Date Posted: 3 years ago
Views: 226
J
J
3 years ago
(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) )
Candice
Candice
3 years ago
Thank you!
J
J
3 years ago
Welcome. The concept used here is partial fractions.

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Sue Xian
Sue Xian's answer
3 answers (Tutor Details)
1st
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
J
J
3 years ago
Slight error in the final answer. The negative sign from '1 - ' applies to both fractions.
J
J
3 years ago
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
Candice
Candice
3 years ago
Thank you!