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.
Appreciate your help,
Alex
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.
Appreciate your help,
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 ?
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 $\displaystyle \pm \sqrt{b^2-4ac} \approx \pm b$
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.