Say if we have:
$\displaystyle x^2 + 12x + 5$
$\displaystyle (x^2 + 6) - 36 + 5$
$\displaystyle (x^2 + 6) - 31$
$\displaystyle x = -6, y = -31$
How how could I find the TP if a != 1?
That should be $\displaystyle (x+6)^2 - 31$.
Anyway, to complete the square when there is a leading coefficient, just factor it out of the first two terms:
$\displaystyle 2x^2 - 12x + 5$
$\displaystyle =2\left(x^2 - 6x\right) + 5$
$\displaystyle =2\left(x^2 - 6x + 9 - 9\right)+5$
$\displaystyle =2\left(x^2 - 6x + 9\right)-2(9)+5$
$\displaystyle =2(x - 3)^2-13$
By the way. How would I go about solving this with logs? 2^x + 2^(x+3) = 36
I know it can be solved simply by breaking 2^(x + 3) into 2^x and 2^3. Giving you 2^x * 8, and then just adding 1 and making it 9.
But there is a way to do it give logarithms is there not?
Right, after you do that you have
$\displaystyle 9\cdot2^x=36$
$\displaystyle \Rightarrow2^x=4$
Now, it's probably obvious that the answer is 2. But we could use logarithms to make our work more explicit:
$\displaystyle 2^x=4\quad\Rightarrow\quad\log_22^x=\log_24 \quad\Rightarrow\quad x=\log_24=2$.