10v^{2}−15v =27+4v^{2}−6v
SOS-
the answer is
v=3 or -3/2
show the calculation please!
$\displaystyle 10v^2 - 15v = 27 + 4v^2 - 6v$
$\displaystyle 6v^2 - 9v = 27$
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
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?
$\displaystyle (v- a)^2= v^2- 2av+ a^2$ 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 $\displaystyle a^2$?