1. ## 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?

2. ## Re: Solving for B

oohh wait a second.. do i use Quadratic equation?

3. ## Re: Solving for B

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 ...

$b = \frac{5 \pm \sqrt{(-5)^2 - 4(5)(1)}}{2(5)}$

4. ## 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...

5. ## Re: Solving for B

just to be sure.. is the answer:

5 +/- sq root of 5 over 10?

(obviously it can be reduced)

6. ## 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.

7. ## 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?

8. ## 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.