# Thread: Rusty on my math

1. ## 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

2. 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:

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

Now, factor out a 3:

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

Now, complete the square in the parenthesis:

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

Now rewrite it as

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

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

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

So it follows that

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

Does this make sense?

3. Hello, Hapa!

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

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

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

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

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

4. ## 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!

5. Originally Posted by Soroban
Hello, Hapa!

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

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

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

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

6. 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.