MX 2 (3 Viewers)

umm what · Apr 21, 2012

is this right?

deswa1 · Apr 21, 2012

Assuming thats (3-2x-x^2) on the denominator, its wrong because you factorised incorrectly.

mirakon · Apr 21, 2012

(3+u)(1-u)

deswa1 · Apr 21, 2012

raufmalik said:
how is that wrong??

(3-2x-x^2)=(x+3)(1-x) not (3-x)(1-x) like you had.

umm what · Apr 21, 2012

I have always problem factorising negative quadratics -.-
I know you pull -1 out (change signs) and then factorise; is that it?

deswa1 · Apr 21, 2012

raufmalik said:
I have always problem factorising negative quadratics -.-
I know you pull -1 out (change signs) and then factorise; is that it?

When you do a few of them, you can do it easily in one step but in the meantime, the way you suggested is good:
3-2x-x^2=-(x^2+2x-3)=-(x+3)(x-1)=(x+3)(1-x)

An easy way to confirm that your answer was wrong was that you had (..-x)(..-x) which would give x^2 when multiplied out not -x^2.

Drongoski · Apr 21, 2012

raufmalik said:
I have always problem factorising negative quadratics -.-
I know you pull -1 out (change signs) and then factorise; is that it?

If you are good at factorising quadratics, you dont have to. But majority of people do as you have indicated, not being confident doing quadratics with a negative x^2 term directly.

$e.g. ~~ 10 - 11x -6x^2 = (2-3x)(5+2x)$

$don't~need: ~ \\ \\ 10-11x-6x^2 = -(6x^2+11x-10) = -(2x+5)(3x-2) = (2-3x)(5+2x)$

umm what · Apr 21, 2012

aha thanks but wjen you change the sign of the roots, how do we decide which root's sign to change?

deswa1 said:
When you do a few of them, you can do it easily in one step but in the meantime, the way you suggested is good:
3-2x-x^2=-(x^2+2x-3)=-(x+3)(x-1)=(x+3)(1-x)

An easy way to confirm that your answer was wrong was that you had (..-x)(..-x) which would give x^2 when multiplied out not -x^2.

deswa1 · Apr 21, 2012

raufmalik said:
aha thanks but wjen you change the sign of the roots, how do we decide which root's sign to change?

Either, it doesn't matter. You could even leave it the way I had it. The reason I changed (x-1) to (1-x) is because that looks a lot neater than (-3-x) which has two negatives.

umm what · Apr 21, 2012

oh alright

thanks mate

deswa1 said:
Either, it doesn't matter. You could even leave it the way I had it. The reason I changed (x-1) to (1-x) is because that looks a lot neater than (-3-x) which has two negatives.

umm what · Apr 21, 2012

Question : Integrate (x^2 + a^2) ^1/2 .dx using integration by parts

:S

nightweaver066 · Apr 21, 2012

raufmalik said:
Question : Integrate (x^2 + a^2) ^1/2 .dx using integration by parts

:S

$\int \sqrt{x^2 + a^2} dx$

$= x\sqrt{x^2 + a^2} - \int \frac{2x}{2\sqrt{x^2 + a^2}} \times xdx$

$= x\sqrt{x^2 + a^2} - \int \frac{x^2}{\sqrt{x^2 + a^2}}dx$

$= x\sqrt{x^2 + a^2} - \int \frac{x^2 + a^2}{\sqrt{x^2 + a^2}} - \frac{a^2}{\sqrt{x^2 + a^2}}dx$

$= x\sqrt{x^2 + a^2} - \int \sqrt{x^2 + a^2} dx + \int \frac{a^2}{\sqrt{x^2 + a^2}}dx$

$\therefore 2\int \sqrt{x^2 + a^2}dx = x\sqrt{x^2 + a^2} + \int \frac{a^2}{\sqrt{x^2 + a^2}}$

$= x\sqrt{x^2 + a^2} + a^2ln(x + \sqrt{x^2 + a^2})$

$\therefore \int \sqrt{x^2 + a^2} dx = \frac{1}{2} (x\sqrt{x^2 + a^2} + a^2ln(x + \sqrt{x^2 + a^2})) + c$

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