1. ## Solving this function with quadratic formula

Hi,

I have been sitting at this function for a couple of hours, trying to solve it.

0.1w^2 + 2155w + 20

I am kind of dumbfounded by this. Even if I factor out the 0.1, I cannot figure this out.

Alex

2. Originally Posted by alex95
Hi,

I have been sitting at this function for a couple of hours, trying to solve it.

0.1w^2 + 2155w + 20

I am kind of dumbfounded by this. Even if I factor out the 0.1, I cannot figure this out.

Alex
$\displaystyle \displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Did you just plug int he numbers?

3. Originally Posted by alex95
Hi,

I have been sitting at this function for a couple of hours, trying to solve it.

0.1w^2 + 2155w + 20

I am kind of dumbfounded by this. Even if I factor out the 0.1, I cannot figure this out.

Alex

$\displaystyle 0.1\left(w^2+21550w\right)+20=0.1\left(w^2+10775w+ 10775w\right)+20$

$\displaystyle =0.1\left(w^2+21550w+10775^2-10775^2\right)+20$

$\displaystyle =0.1\left(w+10775\right)^2-(0.1)10775^2+20$

Are you trying to factor, or is the expression equal to something ?

4. Yes, equal to 200

Thanks,

Alex

5. Now it's equal to something you can put into the form $\displaystyle ax^2+bx+c = 0$ by subtracting 200 from both sides

$\displaystyle 0.1w^2 + 2155w + 20 = 200$ becomes $\displaystyle 0.1w^2+2155w-180 = 0$ then plug and chug.

Of course $\displaystyle 2155^2 >> 4 \cdot 0.1 \cdot -180$ which means $\displaystyle \pm \sqrt{b^2-4ac} \approx \pm b$

6. Ok,

well you can set it up using dwsmith's method, which applies the "abc robot"
but the expression must be equal to zero to use that.

$\displaystyle 0.1w^2+2155w+20=200\Rightarrow\ 0.1w^2+2155w+20-200=200-200=0$

Multiplying zero by anything is zero, so multiply by 10 for convenience

$\displaystyle w^2+21550w-1800=0$

$\displaystyle aw^2+bw+c=0$

$\displaystyle a=1,\;\;\;b=21550,\;\;\;c=-1800$

Alternatively,

$\displaystyle 0.1(w+10775)^2=200-20+(0.1)10775^2$

Multiply both sides by 10.
Take square roots and finally subtract 10775 from both sides.