Eric Nicholas K's answer to LockB's Secondary 4 A Maths Singapore question.
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Q2a
You must practise a lot on quotient rule and the ensuing simplification of expressions.
You must practise a lot on quotient rule and the ensuing simplification of expressions.
Date Posted:
3 years ago
dont really understand what is going on from the 5th line onwards...
This will be hard to explain online because there are lots of things going on in factorisation. It's much easier if this could be done on an interactive whiteboard.
In the fifth line, I merely re-positioned the "1/2" to the front of the numerator.
In the sixth line, I introduced the terms in purple (1/2 and 2) because I wish to factorise a "1/2" out later on. In doing so, I am trying to remove all the "fraction in the numerator or fraction in the denominator".
Also, in the sixth line, I split x^0.5 into x^-0.5 times x^1 because I wish to factorise out x^-0.5 as well.
Remember that common terms will be factorised out in the same way you would have already done in Sec 2, but here it is more difficult because there are powers and other stuff to consider. This is the reason why it's extremely difficult for me to explain it by text where facilities are limited.
Now I have prepared the necessaries for the factorisation step.
In the seventh line, I factored out the common "1/2" and the common "x^-0.5" in the numerator. This leaved behind a "3 + x" for the term before the minus sign and a "2x" after the minus sign in the denominator. I kept the denominator intact.
In the eighth and ninth lines, I simply polished up the numerator to make it look nicer.
Finally, in the tenth line, the "1/2" in the numerator goes down to the denominator as "2" (think why), while the x^-0.5 can be written in positive index form as 1 / x^0.5 in the numerator and thus this goes down to the denominator as x^0.5 in the denominator.
In the fifth line, I merely re-positioned the "1/2" to the front of the numerator.
In the sixth line, I introduced the terms in purple (1/2 and 2) because I wish to factorise a "1/2" out later on. In doing so, I am trying to remove all the "fraction in the numerator or fraction in the denominator".
Also, in the sixth line, I split x^0.5 into x^-0.5 times x^1 because I wish to factorise out x^-0.5 as well.
Remember that common terms will be factorised out in the same way you would have already done in Sec 2, but here it is more difficult because there are powers and other stuff to consider. This is the reason why it's extremely difficult for me to explain it by text where facilities are limited.
Now I have prepared the necessaries for the factorisation step.
In the seventh line, I factored out the common "1/2" and the common "x^-0.5" in the numerator. This leaved behind a "3 + x" for the term before the minus sign and a "2x" after the minus sign in the denominator. I kept the denominator intact.
In the eighth and ninth lines, I simply polished up the numerator to make it look nicer.
Finally, in the tenth line, the "1/2" in the numerator goes down to the denominator as "2" (think why), while the x^-0.5 can be written in positive index form as 1 / x^0.5 in the numerator and thus this goes down to the denominator as x^0.5 in the denominator.
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