1. ## Completing the square.

Original problem: x/12=5/2x+7 (in proportion view) and i got it to: x^2 + 7x=60 and then x^2 + 7/2x = 30 Need help wit the rest please.

2. ## Re: Completing the square!?!?

Originally Posted by livfaure
Original problem: x/12=5/2x+7 (in proportion view) and i got it to: x^2 + 7x=60 and then x^2 + 7/2x = 30 NEED HELP WITH THE REST PLEASE!!
It's not correct! It has to be:
$\frac{x}{12}=\frac{5}{2x}+7 \Leftrightarrow \frac{x^2}{6}=5+14x \Leftrightarrow x^2=30+84x \Leftrightarrow x^2-84x-30=0$

Now you have to reform it to (=completing the square):
$(x-a)^2+b$
where you have to determine a and b.

Can you proceed?

3. ## Re: Completing the square!?!?

No, im sory i guess i wasnt clear enough its like this (see picture) its number 34

4. ## Re: Completing the square!?!?

link to pic is a no show.

x/12 = 5/(2x+7) my parentheses added

$\frac{x}{12} = \frac{5}{2x+7}$

???

5. ## Re: Completing the square!?!?

yes, that it is!

6. ## Re: Completing the square!?!?

Originally Posted by livfaure
yes, that it is!
In that case cross multiply: $x(2x+7) = 60 \Leftrightarrow 2x^2+7x = 60 \Leftrightarrow x^2 + \dfrac{7}{2}x = 30$

The last form is easiest for completing the square. When I complete the square I divide the middle term by 2 (since when we FOIL we get 2ab as the middle term) and add $\frac{b^2}{4}$ (before dividing by 2) to both sides to get the constant term

$x^2 + \dfrac{7}{2}x + \dfrac{49}{16} = 30 + \dfrac{49}{16}$

In terms of the quadratic formula: $x^2 + \dfrac{b}{a} + \dfrac{b^2}{4a^2} = -c + \dfrac{b^2}{4a^2} = \left(x+ \dfrac{b}{2a}\right)^2 = \dfrac{b^2}{4a^2} - c$