Question

solve 3x^2 - 2x -9 =0 by completing the square

Answer 1

please help me :)

Answer 2

to complete the square, you want to arrange your equation to look like
\[x^2 + bx = c\]

I'd start out by adding 9 to each side, and then dividing each side by 3

Answer 3

ok, i did that so i would have x^2-2/3x+3=9 ?

Answer 4

you still there?

Answer 5

[3x^2-2x-9=0]/3 (divide everything by 3)
x^2 - (2/3)x -3 = 0
x^2 -(2/3)x = 3
x^2 - (2/3)x + (1/3)^2 = 3 + (1/3)^2
(x-1/3)^2 = 28/9
sqrt((x-1/3)^2) = +- sqrt(28/9)
x-1/3 = +- (2sqrt(7))/3
x = (1/3)+- (2sqrt(7)/3) or x = [1+- 2sqrt(7)]/3

I hope you get what I wrote.

Answer 6

i mean x^2-2/3x=3...

Answer 7

why did you had (1/3)^2 to each side???

Answer 8

*add

Answer 9

\[x^2 - \frac{2}{3}x =3\]

Now we take half of the coefficient of \(x\), giving us \(-\frac{1}{3}\)

we square that and add to both sides

\[x^2 -\frac{2}{3}x + (-\frac{1}{3})^2 = 3 + (-\frac{1}{3})^2\]

Answer 10

had to add that to complete the square on left side. and whatever you add on left you should do it to right too. (2/3)/2 = 1/3 then you square it to complete the square

Answer 11

now we can rewrite the left side as

\[(x-\frac{1}{3})^2 = 3 + (-\frac{1}{3})^2\]

Answer 12

we have to add the same thing to each side or we break the equation.

\[a = a\]\[a+b = a+b\]

Answer 13

oooohhh I see

Answer 14

so whats the next step?

Answer 15

if we multiply \[(x+a)(x+a) = x^2+ ax + ax + a^2 = x^2 + 2ax + a^2\]you see where the "take half of the coefficient of the x term and square it" comes from.

Answer 16

ok...

Answer 17

if we pretend the coefficient of the x term \(= 2a\) then we can rewrite the left hand side as \((x + a)^2\). that's where the term "completing the square" comes from.

Answer 18

is it x-1/3= 3+/- √-1/3 ???

Answer 19

next will be:
the left side is now a complete square \[x ^{2}-\frac{ 2 }{ 3 }x + (\frac{ 1 }{ 3 })^{2} =\frac{ 28 }{ 9 }\]

Answer 20

wait, how did you get that??

Answer 21

\[(x-\frac{1}{3})^2 = 3 + (-\frac{1}{3})^2\]

\[(x-\frac{1}{3})^2 = 3*\frac{9}9 + (-\frac{1}{3})^2 = \frac{27}{9}+\frac{1}{9} = \frac{28}{9}\]
take square root of both sides:

\[(x-\frac{1}{3}) =\pm \sqrt{\frac{28}{9}} =\pm \frac{2}{3}\sqrt{7}\]

Answer 22

i like this format, thank you!

Answer 23

ok I have a couple ?s... where did you get the 9/9 from?

Answer 24

9/9 = 1, right?

Answer 25

I'm just making a common denominator, because (1/3)^2 = 1/9

Answer 26

yes, but why 9/9

Answer 27

oh

Answer 28

converting 3 to n/9
27/9 = 3, right?

Answer 29

yes, so my final answer is x-1/3= +/- 2/3√7 ?

Answer 30

I'm sorry, this just doesn't make sense? the x has to be by itself right?

Answer 31

move that -1/3 to left side and it will become positive 1/3

Answer 32

you mean the right side?

Answer 33

\[x= \frac{ 1 }{ 3 } \pm \frac{ 2\sqrt{7} }{ 3}\]

Answer 34

in short you are adding 1/3 to both side of equation.

Answer 35

ok, now can u please explain to me how u got the 2√7/3 part?

Answer 36

\[\sqrt{\frac{28}{9}} = \sqrt{\frac{2*2*7}{3*3}} =\frac{2}{3}\sqrt{7}\]

Answer 37

ok: \[x= \frac{ 1 }{ 3 }\pm \frac{ \sqrt{28} }{ 3 }\]
then
\[\sqrt{28} = \sqrt{4\times7}\] right?
so we take out 4 since \[\sqrt{4} = 2\]
results to:
\[\sqrt{4\times7} = 2\sqrt{7}\]
therefore, the equation is:
\[x = \frac{ 1 }{ 3 }\pm \frac{ 2\sqrt{7} }{ 3 }\]

Answer 38

thank you!

Answer 39

I understand... you guys are great! Thank you both :)

Answer 40

would u be able to help me with one more?