# Thread: Completing the square for turning point when a != 1

1. ## Completing the square for turning point when a != 1

Say if we have:

$x^2 + 12x + 5$
$(x^2 + 6) - 36 + 5$
$(x^2 + 6) - 31$
$x = -6, y = -31$

How how could I find the TP if a != 1?

2. ## Re: Completing the square for turning point when a != 1

Originally Posted by SplashDamage
$x^2 + 12x + 5$
$(x^2 + 6) - 36 + 5$
$(x^2 + 6) - 31$
That should be $(x+6)^2 - 31$.

Anyway, to complete the square when there is a leading coefficient, just factor it out of the first two terms:

$2x^2 - 12x + 5$

$=2\left(x^2 - 6x\right) + 5$

$=2\left(x^2 - 6x + 9 - 9\right)+5$

$=2\left(x^2 - 6x + 9\right)-2(9)+5$

$=2(x - 3)^2-13$

3. ## Re: Completing the square for turning point when a != 1

The 9 is from halving the 6 and doing 3^2 isnt it?

However I do not know where you get the -2 from. You lost me on that step.

4. ## Re: Completing the square for turning point when a != 1

Do you know that a(b+ c)= ab+ ac? He had $2(x^2- 6x+ 9- 9)$ and wrote that as $2(x^2- 6x+ 9)+ 2(-9)$.

5. ## Re: Completing the square for turning point when a != 1

Ah that makes much more sense now that you put it that way. Thanks.

6. ## Re: Completing the square for turning point when a != 1

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?

7. ## Re: Completing the square for turning point when a != 1

Originally Posted by SplashDamage
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

$9\cdot2^x=36$

$\Rightarrow2^x=4$

Now, it's probably obvious that the answer is 2. But we could use logarithms to make our work more explicit:

$2^x=4\quad\Rightarrow\quad\log_22^x=\log_24 \quad\Rightarrow\quad x=\log_24=2$.