Show that $$\left( 2x+3\sqrt{x} \right)^{2}$$
Can be written as $$4x ^{2}+P \sqrt{x ^{3}}+9x$$ Where P is a constand to be found

Question

Show that $$\left( 2x+3\sqrt{x} \right)^{2}$$
Can be written as $$4x ^{2}+P \sqrt{x ^{3}}+9x$$ Where P is a constand to be found

hartnn · Answer

either you can foil it, 
(2x+3 root x) * (2x+3 root x)
to get that form,
or you can directly apply the formula $(a+b)^2= a^2+2ab+b^2$

anonymous · Answer

I did the first way (2x+3 root x) * (2x+3 root x)
Which gave me $$4x ^{2}+12x ^{1.5}+9x$$
Now I'm not sure what to do

loser66 · Answer

1.5 = 3/2 when putting it under the root you have sqrt ( x^3)

hartnn · Answer

thats correct,
so your P is 12, right ? 
the co-efficient of the middle term

hartnn · Answer

what exactly was your doubt ? because you were done :P

hartnn · Answer

$\sqrt{x^3}= (x^3)^{(1/2)}=x^{(3/2)}=x^{1.5}$

anonymous · Answer

Well that wasn't so bad, I was expecting to have to do something else to the equation.
Thanks again hartnn.

hartnn · Answer

oh,ok.
welcome ^_^