algebra question (1 Viewer)

s_bilgrami89 · Oct 14, 2011

can someone please show me how to do Q15 Exercise 1E from the cambridge 3u book

MATHSCAPE · Oct 14, 2011

post the question up.

D94 · Oct 15, 2011

$\\\frac{d}{dx} \left ( x\tan^{-1}x \right ) = \tan^{-1}x\ .1 + x. \frac{1}{1+x^{2}} \\ \\\frac{d}{dx} \left ( x\tan^{-1}x \right ) = \tan^{-1}x\ + \frac{x}{1+x^{2}}$
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$\\\frac{d}{dx} \left ( x\tan^{-1}x \right ) = \tan^{-1}x\ + \frac{x}{1+x^{2}} \\ \\\frac{d}{dx} \left ( x\tan^{-1}x \right ) - \frac{x}{1+x^{2}} = \tan^{-1}x \\ \\ \int_{0}^{1} \tan^{-1}x\ dx = \int_{0}^{1} \frac{d}{dx} \left ( x\tan^{-1}x \right ) - \frac{x}{1+x^{2}}\ dx\\ \\=\left [ x\tan^{-1}x \right ]_{0}^{1} - \int_{0}^{1}\frac{x}{1+x^{2}} \ dx \\ \\=\frac{\pi }{4} - \left [ \frac{1}{2}\ln(1+x^{2}) \right ]_{0}^{1} \\=\frac{\pi }{4} - \frac{1}{2}\ln(2)$

s_bilgrami89 · Oct 15, 2011

oh god thats not the question at all ^, sorry bro i'm a gronk of massive proportions i shouldve said year 11 book anyway heres the question sorry it took me ages, i used the equation thing for the first time
$\frac{a^4-b^4}{a^2-2ab+b^2}\times \frac{a^3-b^3}{a^2b+b^3} \times \frac{a^2b-ab^2+b^3}{a^4+a^2b^2+b^4}$

kooliskool · Oct 15, 2011

$\frac{a^{4}-b^{4}}{a^2-2ab+b^2}\times \frac{a^{3}-b^{3}}{a^2 b+b^3}\times \frac{a^{2}b-ab^{2}+b^{3}}{a^4+a^{2}b^{2}+b^4}$
$=\frac{\left (a^{2}-b^{2} \right )\left (a^{2}+b^{2} \right )}{\left ( a-b \right )^2}\times \frac{\left (a-b \right )\left ( a^2+ab+b^2 \right )}{b\left (a^2+b^2 \right )}\times \frac{b\left (a^{2}-ab+b^{2} \right )}{a^4+a^{2}b^{2}+b^4}$

After cancelling out:
$=\frac{\left (a^{2}-b^{2} \right )}{\left ( a-b \right )}\times \frac{\left ( a^2+ab+b^2 \right )}{1}\times \frac{\left (a^{2}-ab+b^{2} \right )}{a^4+a^{2}b^{2}+b^4}$
$=\frac{\left (a-b \right )\left (a+b \right )}{\left ( a-b \right )}\times \frac{\left ( a^2+ab+b^2 \right )}{1}\times \frac{\left (a^{2}-ab+b^{2} \right )}{a^4+2a^{2}b^{2}+b^4-a^2b^2}$
$=\frac{\left (a+b \right )}{1}\times \frac{\left ( a^2+ab+b^2 \right )}{1}\times \frac{\left (a^{2}-ab+b^{2} \right )}{a^4+2a^{2}b^{2}+b^4-a^2b^2}$
$=\frac{\left (a+b \right )}{1}\times \frac{\left ( a^2+b^2+ab \right )}{1}\times \frac{\left (a^{2}+b^{2}-ab \right )}{\left (a^2+b^2 \right )^2-\left (ab \right )^2}$
$=\frac{\left (a+b \right )}{1}\times \frac{\left ( a^2+b^2+ab \right )}{1}\times \frac{\left (a^{2}+b^{2}-ab \right )}{\left ( a^2+b^2+ab \right )\left (a^{2}+b^{2}-ab \right )}$
$=a+b$

hscishard · Oct 15, 2011

kooliskool said:
$\frac{a^{4}-b^{4}}{a^2-2ab+b^2}\times \frac{a^{3}-b^{3}}{a^2 b+b^3}\times \frac{a^{2}b-ab^{2}+b^{3}}{a^4+a^{2}b^{2}+b^4}$
$=\frac{\left (a^{2}-b^{2} \right )\left (a^{2}+b^{2} \right )}{\left ( a-b \right )^2}\times \frac{\left (a-b \right )\left ( a^2+ab+b^2 \right )}{b\left (a^2+b^2 \right )}\times \frac{b\left (a^{2}-ab+b^{2} \right )}{a^4-a^{2}b^{2}+b^4}$

After cancelling out:
$=\frac{\left (a^{2}-b^{2} \right )}{\left ( a-b \right )}\times \frac{\left ( a^2+ab+b^2 \right )}{1}\times \frac{\left (a^{2}-ab+b^{2} \right )}{a^4+a^{2}b^{2}+b^4}$
$=\frac{\left (a-b \right )\left (a+b \right )}{\left ( a-b \right )}\times \frac{\left ( a^2+ab+b^2 \right )}{1}\times \frac{\left (a^{2}-ab+b^{2} \right )}{a^4+2a^{2}b^{2}+b^4-a^2b^2}$
$=\frac{\left (a+b \right )}{1}\times \frac{\left ( a^2+ab+b^2 \right )}{1}\times \frac{\left (a^{2}-ab+b^{2} \right )}{a^4+2a^{2}b^{2}+b^4-a^2b^2}$
$=\frac{\left (a+b \right )}{1}\times \frac{\left ( a^2+b^2+ab \right )}{1}\times \frac{\left (a^{2}+b^{2}-ab \right )}{\left (a^2+b^2 \right )^2-\left (ab \right )^2}$
$=\frac{\left (a+b \right )}{1}\times \frac{\left ( a^2+b^2+ab \right )}{1}\times \frac{\left (a^{2}+b^{2}-ab \right )}{\left ( a^2+b^2+ab \right )\left (a^{2}+b^{2}-ab \right )}$
$=a+b$

loled when I saw this. All that reduced to 3 characters

kooliskool · Oct 15, 2011

ya lols, amazing isn't it? That's algebra for you

s_bilgrami89 · Oct 15, 2011

kooliskool said:
$\frac{a^{4}-b^{4}}{a^2-2ab+b^2}\times \frac{a^{3}-b^{3}}{a^2 b+b^3}\times \frac{a^{2}b-ab^{2}+b^{3}}{a^4+a^{2}b^{2}+b^4}$
$=\frac{\left (a^{2}-b^{2} \right )\left (a^{2}+b^{2} \right )}{\left ( a-b \right )^2}\times \frac{\left (a-b \right )\left ( a^2+ab+b^2 \right )}{b\left (a^2+b^2 \right )}\times \frac{b\left (a^{2}-ab+b^{2} \right )}{a^4-a^{2}b^{2}+b^4}$

After cancelling out:
$=\frac{\left (a^{2}-b^{2} \right )}{\left ( a-b \right )}\times \frac{\left ( a^2+ab+b^2 \right )}{1}\times \frac{\left (a^{2}-ab+b^{2} \right )}{a^4+a^{2}b^{2}+b^4}$
$=\frac{\left (a-b \right )\left (a+b \right )}{\left ( a-b \right )}\times \frac{\left ( a^2+ab+b^2 \right )}{1}\times \frac{\left (a^{2}-ab+b^{2} \right )}{a^4+2a^{2}b^{2}+b^4-a^2b^2}$
$=\frac{\left (a+b \right )}{1}\times \frac{\left ( a^2+ab+b^2 \right )}{1}\times \frac{\left (a^{2}-ab+b^{2} \right )}{a^4+2a^{2}b^{2}+b^4-a^2b^2}$
$=\frac{\left (a+b \right )}{1}\times \frac{\left ( a^2+b^2+ab \right )}{1}\times \frac{\left (a^{2}+b^{2}-ab \right )}{\left (a^2+b^2 \right )^2-\left (ab \right )^2}$
$=\frac{\left (a+b \right )}{1}\times \frac{\left ( a^2+b^2+ab \right )}{1}\times \frac{\left (a^{2}+b^{2}-ab \right )}{\left ( a^2+b^2+ab \right )\left (a^{2}+b^{2}-ab \right )}$
$=a+b$

thanks bro, the part i got stuck was where you change it to 2a^2b^2 and then just shove a minus a^2b^2 on the end
pretty solid stuff there, need to do more of these to get my algebra up to standard for 4u

kooliskool · Oct 15, 2011

yep, that's right, in 4U you need really strong algebra skills, and you need to learn how to go to where the questions wants. So when you got stuck there, you should realise it's a difference of squares at the top, so I manipulated the bottom to get there

.

Good luck in your practice.

D94 · Oct 15, 2011

s_bilgrami89 said:
thanks bro, the part i got stuck was where you change it to 2a^2b^2 and then just shove a minus a^2b^2 on the end
pretty solid stuff there, need to do more of these to get my algebra up to standard for 4u

If you didn't see this in the first place, another way could be to change $a^4 + a^2b^2+b^4$ into $\frac{a^6-b^6}{a^2-b^2}$ and alike for the other ones, and they will eventually cancel off to get you the same answer.

algebra question (1 Viewer)

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