# Solving for B

• Aug 23rd 2011, 04:37 PM
gurrry
Solving for B
I need help solving for B in this equation:

(NOTE: 5b^2 is 5b squared)

5b^2 - 5b +1 = 0

I started by subtracting -1 from both sides getting:

5b^2 - 5b = -1

Then I pulled out 5b from the left side and got:

5b (b - 1) = -1

And now I have no idea what to do. I think I am doing the entire thing wrong. Can someone help me out?
• Aug 23rd 2011, 04:50 PM
gurrry
Re: Solving for B
oohh wait a second.. do i use Quadratic equation?
• Aug 23rd 2011, 04:51 PM
skeeter
Re: Solving for B
Quote:

Originally Posted by gurrry
I need help solving for B in this equation:

(NOTE: 5b^2 is 5b squared)

5b^2 - 5b +1 = 0

I started by subtracting -1 from both sides getting:

5b^2 - 5b = -1

Then I pulled out 5b from the left side and got:

5b (b - 1) = -1

And now I have no idea what to do. I think I am doing the entire thing wrong. Can someone help me out?

yes, you are.

$5b^2 - 5b + 1 = 0$

the trinomial in not factorable, so use quadratic formula ...

$b = \frac{5 \pm \sqrt{(-5)^2 - 4(5)(1)}}{2(5)}$
• Aug 23rd 2011, 04:53 PM
gurrry
Re: Solving for B
Heh.. I guess I shouldve thought harder before posting. It didnt occur to me to use QF. I guess im still rubbing off the summer "rust". Lol.

thanks for the help! ill more than likely be back sometime today...
• Aug 23rd 2011, 04:58 PM
gurrry
Re: Solving for B
just to be sure.. is the answer:

5 +/- sq root of 5 over 10?

(obviously it can be reduced)
• Aug 23rd 2011, 06:08 PM
HallsofIvy
Re: Solving for B
Another way- which I personally prefer to just plugging into a formula: complete the square:
5b^2- 5b= -1 as you say. Divide through by 5:
b^2- b= -1/5.

(b- a)^2= b^2- 2ab+ a^2 will match that is 2a= 1 so a= -1/2 and a^2= 1/4.

We can make b^2- b into a "perfect square by adding 1/4:
b^2- b+ 1/4= -1/5+ 1/4= -4/20+ 5/20= 1/20
(b- 1/2)^2= 1/20= 1/(4(5))
$b- 1/2= \pm 1/(2(sqrt(5))$

$b= 1/2\pm sqrt(5)/10$ as you say.
• Aug 24th 2011, 03:17 PM
gurrry
Re: Solving for B
Sorry to bump this again, but I have a question reguarding the reducing.

Is the fully reduced answer 1 +/- Sqrt of 1 over 2? Or can whats inside the sq. root not be reduced?
• Aug 24th 2011, 03:30 PM
skeeter
Re: Solving for B
$\frac{5 \pm \sqrt{5}}{10}$

or

$\frac{1}{2} \pm \frac{\sqrt{5}}{10}$

the first one is considered "reduced" as is ... the second is just a splitting of the fractions.