Rusty on my math

• Aug 27th 2009, 07:15 PM
Hapa
Rusty on my math
If someone could step me through how to do the following equation again, it would be greatly appreciated. Hard for me to get back into the swing of things again, LOL!! The equation is as follows:

If 3x^2 - 2x +7 = 0, then (x - 1/3)^2 = ?

Thanks again!

v/r
Hapa
• Aug 27th 2009, 07:26 PM
Chris L T521
Quote:

Originally Posted by Hapa
If someone could step me through how to do the following equation again, it would be greatly appreciated. Hard for me to get back into the swing of things again, LOL!! The equation is as follows:

If 3x^2 - 2x +7 = 0, then (x - 1/3)^2 = ?

Thanks again!

v/r
Hapa

First, group the first two terms together in the first equation:

$(3x^2-2x)+7=0$.

Now, factor out a 3:

$3\left(x^2-\tfrac{2}{3}x\right)+7=0$

Now, complete the square in the parenthesis:

$3\left(x^2-\tfrac{2}{3}x+\tfrac{1}{9}-\tfrac{1}{9}\right)+7=0$

Now rewrite it as

$3\left(x^2-\tfrac{2}{3}+\tfrac{1}{9}\right)+7-\tfrac{1}{3}=0$

So it follows that the terms in the parenthesis form a perfect square:

$3\left(x-\tfrac{1}{3}\right)^2+\tfrac{20}{3}=0$.

So it follows that

$\left(x-\tfrac{1}{3}\right)^2=\color{red}\boxed{-\tfrac{20}{9}}$.

Does this make sense?
• Aug 27th 2009, 07:33 PM
Soroban
Hello, Hapa!

Quote:

If $3x^2 - 2x +7 \:=\: 0$, then $\left(x -
\frac{1}{3}\right)^2 \:=\: ?$

We have: . $3x^2 - 2x + 7 \:=\:0 \quad\Rightarrow\quad 3x^2 - 2x \:=\:-7$

Divide by 3: . $x^2 - \frac{2}{3}x \:=\:-\frac{7}{3}$

Add $\frac{1}{9}\!:\quad x^2 -\frac{2}{3}x + {\color{blue}\frac{1}{9}} \:=\:-\frac{7}{9} + {\color{blue}\frac{1}{9}}$

Therefore: . $\left(x - \frac{1}{3}\right)^2 \;=\;-\frac{4}{9}$

• Aug 27th 2009, 07:40 PM
Hapa
Different solution
It is starting to come back to me...yet the solution that was given for this equation was :

(x - 1/3)^2 = -20/9

How did they get that?

Thanks again!
• Aug 27th 2009, 07:42 PM
Chris L T521
Quote:

Originally Posted by Soroban
Hello, Hapa!

We have: . $3x^2 - 2x + 7 \:=\:0 \quad\Rightarrow\quad 3x^2 - 2x \:=\:-7$

Divide by 3: . $x^2 - \frac{2}{3}x \:=\:-\frac{7}{3}$

Add $\frac{1}{9}\!:\quad x^2 -\frac{2}{3}x + {\color{blue}\frac{1}{9}} \:=\:-\frac{7}{{\color{red}9}} + {\color{blue}\frac{1}{9}}$

Therefore: . $\left(x - \frac{1}{3}\right)^2 \;=\;-\frac{4}{9}$

I think you made a minor error. The part in red should be 3, and then your result will lead to the value of my corrected answer.
• Aug 27th 2009, 07:43 PM
Chris L T521
Quote:

Originally Posted by Hapa
It is starting to come back to me...yet the solution that was given for this equation was :

(x - 1/3)^2 = -20/9

How did they get that?

Thanks again!

It is -20/9. Please see my corrected post.