1. ## Solve an equation

Solve
\$\displaystyle \frac{3}{(y-1)^2}+\frac{3}{2(y-1)^2}=2$

I tried most things but get stuck at $\displaystyle \frac{9}{4}=(y-1)^2$ pretty much every time. Any help?

2. Originally Posted by Mukilab
Solve
$\displaystyle \frac{3}{(y-1)^2}+\frac{3}{2(y-1)^2}=2$

I tried most things but get stuck at $\displaystyle \frac{9}{4}=(y-1)^2$ pretty much every time. Any help?
You're fine up to there, you can complete the square to find y. Remember that $\displaystyle \frac{9}{4} = \left(\frac{3}{2}\right)^2$

3. How can I complete the square with that!?

I thought completing the square is along the lines of

2x^2+4x+13=(2x^2+4x+4)-11

o_o

What can I do with {y-1}^2?

y^2-2y+1

completing the square

(y^2-2y+1)+1-1=9/4

4. Originally Posted by Mukilab
How can I complete the square with that!?

I thought completing the square is along the lines of

2x^2+4x+13=(2x^2+4x+4)-11

o_o

What can I do with {y-1}^2?

y^2-2y+1

completing the square

(y^2-2y+1)+1-1=9/4
$\displaystyle (y-1)^2$ is already in completed square form so you can take the square root. In other words you can take the square root of both sides and then add 1.

5. Originally Posted by e^(i*pi)
$\displaystyle (y-1)^2$ is already in completed square form so you can take the square root. In other words you can take the square root of both sides and then add 1.
I've done that already but I got 2 and 4 which don't work if you put them into the equation

6. OOhhhhh

Thanks, I forgot when I did it you have to take the square root on both sides! heh.

Don't I need an exact answer though? Not allowed a calculate

final:
$\displaystyle y=\sqrt{\frac{9}{4}}+1$

7. Originally Posted by e^(i*pi)
Remember that $\displaystyle \frac{9}{4} = \left(\frac{3}{2}\right)^2$
Originally Posted by Mukilab
OOhhhhh

Thanks, I forgot when I did it you have to take the square root on both sides! heh.

Don't I need an exact answer though? Not allowed a calculate

final:
$\displaystyle y=\sqrt{\frac{9}{4}}+1$
See post 2 (quoted at the top of this post) for how to eliminate the square root. Also you still need the $\displaystyle \pm$ sign.

edit: that answer is correct, it just needs simplifying and also the negative solution

8. Originally Posted by e^(i*pi)
See post 2 (quoted at the top of this post) for how to eliminate the square root. Also you still need the $\displaystyle \pm$ sign.

edit: that answer is correct, it just needs simplifying and also the negative solution
$\displaystyle y=\pm\sqrt{\frac{13}{4}}$????

sorry

9. Originally Posted by Mukilab
$\displaystyle y=\sqrt{\frac{9}{4}}+1$
That's 3/2 + 1 = 5/2

10. Originally Posted by Mukilab
$\displaystyle y=\pm\sqrt{\frac{13}{4}}$????

sorry
Don't ever be sorry for asking for help

$\displaystyle \pm \sqrt{\frac{9}{4}} = \pm \sqrt{\left(\frac{3}{2}\right)^2}$

As $\displaystyle \pm \sqrt{a^2} = \pm a$ that square root can be reduced

You should get $\displaystyle y = \pm \frac{3}{2} + 1$ hence $\displaystyle y = \frac{5}{2} \text{ or } y = -\frac{1}{2}$

We can't add the 1 under the square root since $\displaystyle \sqrt{a} + \sqrt{b} \neq \sqrt{a+b}$

11. Originally Posted by e^(i*pi)
Don't ever be sorry for asking for help

$\displaystyle \pm \sqrt{\frac{9}{4}} = \pm \sqrt{\left(\frac{3}{2}\right)^2}$

As $\displaystyle \pm \sqrt{a^2} = \pm a$ that square root can be reduced

You should get $\displaystyle y = \pm \frac{3}{2} + 1$ hence $\displaystyle y = \frac{5}{2} \text{ or } y = -\frac{1}{2}$

We can't add the 1 under the square root since $\displaystyle \sqrt{a} + \sqrt{b} \neq \sqrt{a+b}$