1. ## Complete the square

10v2−15v =27+4v2−6v
SOS-
v=3 or -3/2

2. ## Re: Complete the square

Originally Posted by edmundyong
10v2−15v =27+4v2−6v
SOS-
v=3 or -3/2

$\displaystyle 10v^2 - 15v = 27 + 4v^2 - 6v$

$\displaystyle 6v^2 - 9v = 27$

I'm sure you've gotten this far.

What does "complete the square" mean? We need to add a number to both sides of the equation such that the LHS is a perfect square. My first suggestion is to divide through by 6 to make the coefficient of v^2 one. (This is not necessary, but I find it makes the problem simpler.)

$\displaystyle v^2 - \frac{9}{6}v = \frac{27}{6}$

$\displaystyle v^2 - \frac{3}{2}v = \frac{9}{2}$

We need to find a number to add to the LHS to make it a perfect square. The only number we can use is the coefficient of the linear term: 3/2. What does that tell us about the number we must add? Check your text and see how far you can get from here.

-Dan

3. ## Re: Complete the square

Here is another approach to completing the square, arrange the equation as:

$\displaystyle 2v^2-3v=9$

Multiply through by 4 times the coefficient of the squared term, which is 8 and then add the square of the coefficient of the linear term which is 9:

$\displaystyle 16v^2-24v+9=72+9$

$\displaystyle (4v-3)^2=9^2$

Can you finish?

4. ## Re: Complete the square

$\displaystyle (v- a)^2= v^2- 2av+ a^2$ You can write $\displaystyle 10v^2−15v =27+4v^2−6v$ as $\displaystyle 10v^2- 15v- (4v^2- 6v)= 6v^2- 9v= 6(v^2- (3/2)v= 27$.

Now compare $\displaystyle v^2- (2/3)v$ with $\displaystyle v^2- 2av$. What is "a"? What is $\displaystyle a^2$?