Question

solve the quadratic equation by completing the square. -3x^2+9x=1

Answer 1

for the equation \(ax^2+bx+c=0\) we can complete the square with this
\(a(x+\frac{b}{2a})^2=-c+a(\frac{b}{2a})^2\)

Answer 2

right..........but i did it in the classic way......jus got stuck in the options though...........so far i solved up till here......... -3x^2 +9x +81/4 - 1 - 81/4=0

Answer 3

confused bro..? i did the (b/2)^2

Answer 4

that gave me 81/4

Answer 5

\[
-3x^2+9x = -3(\color{red}{x^2}\color{blue}{-3}\color{red}{x})
\]Now we use \[
(x+a)^2 = x^2+2a+a^2 \\
(\color{purple}{x+a})^2-\color{green}{a^2}=\color{red}{x^2}\color{blue}{+2a}\color{red}{x}
\]We let \(-3=2a\) and get \(a=-3/2\) and \(a^2 = 9/4\).\[
x^2-3x = (\color{purple}{x-3/2})^2-\color{green}{9/4}
\]

Answer 6

We sub that back in \[
-3x^2+9x = -3[(x-3/2)^2-9/4] = -3(x-3/2)^2 - 27/4
\]

Answer 7

Actually, probably should have stuck with\[
(x-3/2)^2-9/4 = -\frac 13
\]

Answer 8

I think you can do the rest.

Answer 9

yea.....i see........but take a look at this please

Answer 10

And?

Answer 11

You have to finish the rest of it.

Answer 12

You're not sure what to do next?

Answer 13

thats the problem dude.......i end up using calculator and dividing everything resulting in decimals

Answer 14

Do you know how to add fractions? Don't use a calculator.

Answer 15

yes i do

Answer 16

i tried ............2 full pages sitting in trash.....so i came here

Answer 17

please help me man..............my brain is frozen......blank!?!?!

Answer 18

So you want me to teach you how to add fractions?

Answer 19

\[
(x-3/2)^2 = -\frac 13 + \frac 94
\]

Answer 20

lol......no wait

Answer 21

x - 3/2 = sq.root of 23/12........right?

Answer 22

k wat next now.....?

Answer 23

uhh @wio ??

Answer 24

well it is \[
|x-3/2| =\sqrt{\frac{23}{12}}
\]

Answer 25

\[
x=\frac 32 \pm \sqrt{\frac{23}{12}}
\]

Answer 26

yea.....exactly wat i said.....wat next though?

Answer 27

the answer choices dont quite match...........the sq.root portion.....

Answer 28

@wio ........u there buddy?

Answer 29

@zzr0ck3r .......u there?

Answer 30

hold on

Answer 31

k sorry.........m hasty.....lol

Answer 32

lol

Answer 33

@zzr0ck3r .........we are stuck here....nd u think it's funny?

Answer 34

wio aint stuck, im guessing hes busy

Answer 35

dont think so......he got the first part right........but the 2nd one....he might be working on i guess?

Answer 36

\( -3x^2+9x=1\implies 3x^2-9x=-1\implies3x^2-9x+1=0
\)

so

\(3(x^2-3x)+1=0\\3(x-\frac{3}{2})^2+1=3(\frac{3}{2})^2\\3(x-\frac{3}{2})^2=\frac{27}{4}-1=\frac{23}{4}\)

so

\((x-\frac{3}{2})^2=\frac{23}{12}\implies x-\frac{3}{2}=\pm\sqrt{\frac{23}{12}}\\x=\frac{3}{2}\pm \sqrt{\frac{23}{12}}\)

Answer 37

@zzr0ck3r ..... we got there buddy.........but take a look at this attachment please....

Answer 38

D

Answer 39

@zzr0ck3r ........so what do you think..??.............

Answer 40

\(x=\frac{3}{2}\pm \sqrt{\frac{23}{12}}\rightarrow\frac{3}{2}\pm\frac{\sqrt{3}}{\sqrt{3}}\sqrt{\frac{23}{12}}=\frac{3}{2}\pm\sqrt{\frac{69}{36}}=\frac{3}{2}\pm\frac{\sqrt{69}}{6}\)

Answer 41

so D is the answer

Answer 42

\[\sqrt{\frac{ 23 }{ 12 }(\frac{ 3 }{ 3 })} = \sqrt{\frac{ 69 }{ 36 }} = \frac{ \sqrt{69} }{ 6 }\]

Answer 43

\(x=\frac{3}{2}\pm \sqrt{\frac{23}{12}}\rightarrow\frac{3}{2}\pm1*\sqrt{\frac{23}{12}}=\frac{3}{2}\pm\frac{\sqrt{3}}{\sqrt{3}}\sqrt{\frac{23}{12}}=\frac{3}{2}\pm\sqrt{\frac{69}{36}}=\frac{3}{2}\pm\frac{\sqrt{69}}{6}\)

Answer 44

ohh wow.........@zzr0ck3r and @Johnbc .....thanks guyz....nd @wio too..!!

Answer 45

so how did you get 3/3.....?

Answer 46

look at my steps

Answer 47

1 = sqrt(3)/sqrt(3)

Answer 48

when we move that in the sqrt we get 3/3

Answer 49

Its just multiplying the square root fraction by one as zzr0ck3r demonstrated

Answer 50

ahh...........i see......i didnt know u could substitute something in a sq.root like this....

Answer 51

\(\frac{3}{2}\pm\color{blue}{1}*\sqrt{\frac{23}{12}}=\frac{3}{2}\pm\color{blue}{\frac{\sqrt{3}}{\sqrt{3}}}\sqrt{\frac{23}{12}}=\frac{3}{2}\pm\sqrt{\frac{23*\color{blue}{3}}{12*\color{blue}{3}}}=\frac{3}{2}\pm\frac{\sqrt{69}}{6}\)

Answer 52

yeah \(\sqrt{a}*\sqrt{b} = \sqrt{ab}\)

Answer 53

but \(\sqrt{a}+\sqrt{b}\ne \sqrt{a+b}\)

Answer 54

in general....

Answer 55

k .... @zzr0ck3r ........ thx a lot man........ nd for the extra hints ....... appreciate it :D

Answer 56

np, I guarantee you that wio was just busy:)

Answer 57

lol........ seems like u knw him well