# Math Help - Alg. 2, Quadratics

I am having trouble with two problems.

1.

Edit: figured it out

2.

3(x+7/3)^2 = 1

I'm just lost on that one D:

2. Originally Posted by ajm490
I am having trouble with two problems.

1.

(x-3/5)^2 = 4/5

I turn it into x^2 -6/5X+9/25= 4/5 and then x^2 -6/5X-11/25 = 0.

Then I get stuck trying to plug that into the quadratic formula.
$
(x-3/5)^2 = 4/5
$

so:

$
x-3/5 = \pm \sqrt{4/5}
$

hence:

$
x = 3/5 \pm \sqrt{4/5}
$

No need for the quadratic formula

RonL

3. Originally Posted by ajm490
2.

3(x+7/3)^2 = 1
do not delete your questions after they have been solved

$3(x + 7/3)^2 - 1 = 0$

$\Rightarrow 3 \left( x^2 + \frac {14}3x + \frac {49}{9} \right) - 1 = 0$

can you take it from here?

4. Yes, I think so. And I found the answer by myself actually, prior to him completing it, which is why I took it off.

Thanks to both of you though.

5. Originally Posted by ajm490
I am having trouble with two problems.

1.

Edit: figured it out

2.

3(x+7/3)^2 = 1

I'm just lost on that one D:

Same as other one:

$
3(x+7/3)^2 = 1
$

so:

$
(x+7/3)^2 = 1/3
$

then:

$
x+7/3 = \pm \sqrt{1/3}
$

hence:

$
x = -7/8 \pm \sqrt{1/3}
$

RonL

6. Originally Posted by CaptainBlack
Same as other one:

$
3(x+7/3)^2 = 1
$

so:

$
(x+7/3)^2 = 1/3
$

then:

$
x+7/3 = \pm \sqrt{1/3}
$

hence:

$
x = -7/8 \pm \sqrt{1/3}
$

RonL
i was working too hard

7. Originally Posted by Jhevon
i was working too hard
I had thought of mntioning that, but then ..

RonL