Question

find the value b in each perfect square trinomial:
3x^2+bx+27 explain please.

Answer 1

please help

Answer 2

perfect square are those you can write as
\[(tx \pm ry)^2\]
When you expand a binomial in tthat form you'll get
\[t^2x^2 \pm 2txry + r^2y^2\]

In your case, \[r^2y^2=27\rightarrow ry=3\sqrt {3}\]
\[t^2x^2=3x^2 \rightarrow tx=\sqrt {3} x\]
Then
\[2txry= 18x \]
In your case "b"=2try
So, you'll get b=18
Got it?

Answer 3

i got how u got 3√3
but then no
what did u exactly do?

Answer 4

The exact same, aply a square root to t^2x^2, to obtain tx

Answer 5

so 3√3 multiply by what?

Answer 6

by tx, the easiest explanation says
\[(ax+c)^2=a^2+2acx+c^2\], in that case, your "b" will be 2ac, got it?

Answer 7

yes so first we divid everything by 3?

Answer 8

if you do that, you'll get b/3

Answer 9

yes is that correct?

Answer 10

hmm...show me how you do

Answer 11

i dont get it :'(

Answer 12

if i divid by 3 ill get x^2+bx+9
i take c multiply it by 2 :D

Answer 13

you'll get x^2+(b/3)x+9

Answer 14

i dont get your way its so complicated :/ where did you get tryx from

Answer 15

random constants xD

Answer 16

lol choose x lol