Question

If\[3x^{2}-2x+7=0\]then\[\left(x-\frac{1}{3}\right)^{2}=\]My thoughts:
1. solve using quadratic formula
2. input x into the second equation

Just checking my work here.

Answer 1

That would work..can you plug it into the quadratic formula?

Answer 2

This is the formula yes?\[x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\]

Answer 3

So -\[x=\frac{-(-2)\pm\sqrt{(-2)^{2}-4(3)(7)}}{2(3)}\]

Answer 4

Am I on the right track here?

Answer 5

What do you think? @hartnn

Answer 6

try another method which is called squaring.

Answer 7

squaring?

Answer 8

Your quadratic formula set-up is correct.

Answer 9

yep squaring ,not familiar ?

Answer 10

I'm only in high school ^^;
Thank you @iGreen.

Answer 11

\[=\frac{2\pm\sqrt{4-84}}{6}=\frac{2\pm\sqrt{-80}}{6}=\frac{2\pm i\sqrt{80}}{6}\]

Answer 12

Any errors yet? :o

Answer 13

\(\large \begin{align} \color{black}{3x^2-2x+7=0\hspace{.33em}\\~\\
x^2-2\cdot\dfrac{1}{3}x+\dfrac{7}{3}=0\hspace{.33em}\\~\\
x^2-2\cdot\dfrac{1}{3}x+\left(\dfrac{1}{3}\right)^2-\left(\dfrac{1}{3}\right)^2+\dfrac{7}{3}=0\hspace{.33em}\\~\\
}\end{align}\)

Answer 14

. . .
# too complex
# teacher never showed us that
# no thanks lol

Answer 15

(but yes, I understand the gist of it)

Answer 16

ok !, was easy though

Answer 17

Well it would take more time on my computer-based exam tomorrow. I'd rather use a formula I'm familiar with -

Answer 18

(These are NOT exam questions!)

Answer 19

Have I made any errors so far in my quadratic equation?

Answer 20

by the my method was based on this

\(\Large (a-b)^2=a^2-2ab+b^2\)

Answer 21

\[\text{subsize }\sqrt{80}\text{ becomes }4\sqrt{5}\]\[\text{so: }4i\sqrt{5}\]\[\text{which becomes: }\frac{2\pm 4i\sqrt{5}}{6}\]

Answer 22

I think I misused the word "subsize"

Answer 23

...and that's as far as I got really

Answer 24

@iGreen. did I make any errors?

Answer 25

Looks correct..

Answer 26

But the problem is, we can't find the square root of a negative number..I'm stumped. @iambatman

Answer 27

Can I go any further ...

I guess not

Answer 28

\(\Large x =\frac{2\pm i\sqrt{80}}{6}= \frac{1}{3} \pm i\frac{\sqrt{80} }{6} \\ \Large x -\dfrac{1}{3} = \pm i\frac{\sqrt{80} }{6} \)

ok till here ?
bdw, mathmath's way was easier, but lets go with whatever you are comfortable with :)

Answer 29

and we don't have to find the square root now, we have to find the square, and square of imaginary number is easy :)
\(\i^2 =-1\)

Answer 30

wha-
ok-

Answer 31

and you just have to square both sides of this :
\(\Large x =\frac{2\pm i\sqrt{80}}{6}= \frac{1}{3} \pm i\frac{\sqrt{80} }{6} \\ \Large x -\dfrac{1}{3} = \pm i\frac{\sqrt{80} }{6}\)
:)

Answer 32

@hartnn I was taught that you cannot simplify out half of an addition/subtraction equation

Answer 33

whats a
"half of addition equation??"

Answer 34

like \[\frac{8+5}{2}\] becomes\[\frac{4+5}{1}\]

Answer 35

I meant \[\sqrt{5}\] sorry

Answer 36

yes, that is indded incorrect
but thats not what i did

Answer 37

^

Answer 38

^^
is perfectly valid