Question

Help Please

f(x) = 3x2 + 18x – 5 put in standard form by the method of completing the square

Answer 1

\[3x^2 + 18x – 5=0\]\[3x^2 + 18x = 5\]\[3x^2 + 18x \color{red} {+27}= 5\color{red} {+27}\]\[3x^2+18x+27=32\]\[\color{red} {3}(x^2+6x+9)=32\]\[x^2+6x+9=32/3\]\[(x+3)^2=32/3\]

Answer 2

wait how'd you get 27 half of 18 is 9 then 9^2 is 81?

Answer 3

\[x+3=±~\sqrt{ \frac{32}{3} }\]\[x+3=±~ \frac{\sqrt{32}}{\sqrt{3}}\]\[x+3=±\frac{\sqrt{96}}{3}\]\[x=-3±\frac{\sqrt{96}}{3}\]

Answer 4

\[x=\frac{-9}{3}±\frac{\sqrt{96}}{3}\]

Answer 5

\[x=\frac{-9± \sqrt{96}}{3}\]\[x=\frac{-9±4\sqrt{6}}{3}\]

Answer 6

You lost me I didn't have to do all that during class

Answer 7

Where did I lose you?

Answer 8

3x2+18x+27=5+27

Answer 9

aren't you suppose to take half of 18 then square it

Answer 10

I am supposed to make the left side into a perfect square and that's what I did.

Answer 11

but when you square 9 I get 81

Answer 12

I know the vertex is (-3.-32)

Answer 13

Do you get how I solved for x?

Answer 14

not completely