$\displaystyle x^2-7x+12=0$

$\displaystyle 2x^2 - 7x - 4 = 0$

Use the zero product properity to solve for x and k

$\displaystyle 6k^2-k-1=0$
$\displaystyle x^2+6x-7=0$

I hate posting this many problems but the math class i'm in, I can't learn, its an overcrowded classroom with people who are there that don't care to learn and just do whatever, I'm getting my mom to call so I can get switched out of the class and put into one that i'll have the ability to learn more, anyways, thank you for any help that anyone is able to give.

2. Originally Posted by Haxcake

$\displaystyle x^2-7x+12=0$

$\displaystyle 2x^2 - 7x - 4 = 0$
Given $\displaystyle ax^2 + bx + c = 0$ then $\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

So for the first one:
$\displaystyle x^2-7x+12=0$

a = 1, b = -7, c = 12

Thus:
$\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$\displaystyle x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(12)}}{2 \cdot 1}$

$\displaystyle x = \frac{7 \pm \sqrt{49 - 48}}{2}$

$\displaystyle x = \frac{7 \pm \sqrt{1}}{2}$

$\displaystyle x = \frac{7 \pm 1}{2}$

So $\displaystyle x = \frac{7+1}{2} = 4$ or $\displaystyle x = \frac{7 - 1}{2} = 3$.

I'll let you do the second one. I got x = 4 or x = -1/2.

-Dan

3. Originally Posted by Haxcake
Use the zero product properity to solve for x and k

$\displaystyle 6k^2-k-1=0$
$\displaystyle x^2+6x-7=0$
This uses the property of real numbers:
If ab = 0 then either a = 0 or b = 0 (or both).

The rest of it is factoring.

The first one says:
$\displaystyle 6k^2-k-1=0$

$\displaystyle (2k - 1)(3k + 1) = 0$ (If you need help with factoring, just let me know.

So either $\displaystyle 2k - 1 = 0$ ==> $\displaystyle k = 1/2$
or
$\displaystyle 3k + 1 = 0$ ==> $\displaystyle k = -1/3$.

Again, I'll leave the other one to you. I get x = 1 or x = -7.

-Dan

4. The quadratic formula can be looked up anywhere.

As for the second half of your post:

I reckon by 'zero product property', they mean factoring?. If so, why don't they just say so.
If I am correct, why would they call it that except to show off and act pretentious.

Factor:

$\displaystyle 6k^{2}-k-1$

Because the leading coefficient is 6, we must compensate and find

what numbers add to -1 and when multiplied equal -6.

That would be 2 and -3, wouldn't it?.

$\displaystyle 6k^{2}+2k-3k-1$

$\displaystyle (6k^{2}+2k)-(3k+1)$

Factor:

$\displaystyle 2k(3k+1)-(3k+1)$

$\displaystyle (2k-1)(3k+1)$

I hope this is what they mean by zero product property.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\displaystyle x^{2}+6x-7$

Do the same thing:

What 2 numbers when added equal 6 and when multiplied equal -7

That'd be 7 and -1.

$\displaystyle x^{2}-x+7x-7$

$\displaystyle (x^{2}-x)+(7x-7)$

Factor:

$\displaystyle x(x-1)+7(x-1)$

$\displaystyle (x-1)(x+7)$

5. ## i am doin the same thing i could only get the first one and the last on tough.

Originally Posted by Haxcake

$\displaystyle x^2-7x+12=0$

$\displaystyle 2x^2 - 7x - 4 = 0$

Use the zero product properity to solve for x and k

$\displaystyle 6k^2-k-1=0$
$\displaystyle x^2+6x-7=0$

I hate posting this many problems but the math class i'm in, I can't learn, its an overcrowded classroom with people who are there that don't care to learn and just do whatever, I'm getting my mom to call so I can get switched out of the class and put into one that i'll have the ability to learn more, anyways, thank you for any help that anyone is able to give.
1. $\displaystyle x^2-7x+12=0$
(x-3)(x-4)
x-3=0 x=4
x-4=0 x=3

4. $\displaystyle x^2+6x-7=0$
(x+7)(x-1)
x+7=0 x=1
x-1=0 x=-7