# Math Help - The Quadratic Formula

Using the quadratic formula, find the exact value of x for the following.

-2kx^2 + 4x + 6k = 0

2. Originally Posted by Joker37
Using the quadratic formula, find the exact value of x for the following.

-2kx^2 + 4x + 6k = 0
The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

In your case - assuming k is a constant,

$
a = -2k
b = 4
c = 6k
$
.

Just put these into your answer and solve for x. Your answer will have $k$ in it, the "exact value" will most probably involve surds.

Hope this helps

3. Originally Posted by Joker37
Using the quadratic formula, find the exact value of x for the following.

-2kx^2 + 4x + 6k = 0
If you have an equation that says $ax^2+bx+c=0$ then the quadratic formula is as follows (for $k \neq f(x)$):

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

• a is whatever is in front of x^2 = -2k
• b is whatever is in front of x = 4
• c is whatever does not have an x with it = 6k

Thus:

$x = \frac{-4 \pm \sqrt{16- (4 \times -2k \times 6k)}}{-4k}$

Simplify to give $x = \frac{1 \mp \sqrt {3k^2+4}}{k}$

There is no way to solve x for a number without knowing the value of k

4. How did you get from this...

Originally Posted by e^(i*pi)

Thus:

$x = \frac{-4 \pm \sqrt{16- (4 \times -2k \times 6k)}}{-4k}$
...to this.

Originally Posted by e^(i*pi)
Simplify to give $x = \frac{1 \mp \sqrt {3k^2+4}}{k}$

5. Originally Posted by Joker37
How did you get from this...

...to this.

$x = \frac{-4 \pm \sqrt{16- (4 \times -2k \times 6k)}}{-4k}$

Collecting and multiplying out the terms in the square root gives $16-(-48k^2) = 48k^2+16$

As the HCF of 48 and 16 is 16 we can remove this as a factor: $48k^2+16 = 16(3k^2+1)$ note, I made a mistake here in my other working - that +3 should have been a +1

Reintroducing this to the full equation:

$\frac{-4 \pm \sqrt{16(3k^2+1)}}{-4k}$

As $\sqrt{ab} = \sqrt{a} \sqrt{b}$ then $\sqrt{16(3k^2+1)} = \sqrt{16} \times \sqrt{3k^2+1} = 4\sqrt{3k^2+1}$

Reintroducing the simplified discriminant:

$\frac{-4 \pm 4\sqrt{3k^2+1}}{-4k} = \frac{4(-1 \pm \sqrt{3k^2+1})}{-4k}$

4 will cancel to give

$\frac{-1 \pm \sqrt{3k^2+1}}{-k}$

You can leave it like that but it's more usual to not have a negative denominator so we multiply by $\frac{-1}{-1}$

$-\frac{-1 \pm \sqrt{3k^2+1}}{-k} = \frac{1 \mp \sqrt{3k^2+1}}{k}$

Since there is a k² term under the square root sign then x will be real for all real k

6. Just to check, $\pm$ and $\mp$ are both the same things, right?

(This is my first post where I have used LaTex , sort of exciting in a way!).

7. Originally Posted by Joker37
Just to check, $\pm$ and $\mp$ are both the same things, right?

(This is my first post where I have used LaTex , sort of exciting in a way!).
The same numbers but in different order!

Notice that he had first $\frac{-1\pm\sqrt{3k^2+1}}{-k}$ which gives the two solutions $\frac{-1+\sqrt{3k^2+1}}{-k}= \frac{1-\sqrt{3k^2+1}}{k}$ and $\frac{-1- \sqrt{3k^2+1}}{-k}= \frac{1+\sqrt{k^2+1}}{k}$.

e^(i*pi) just changed from $\pm$ to $\mp$ to make it clear that the $-\sqrt{k^2+1}$ had changed to $+\sqrt{k^2+1}$ and vice-versa. Of course, it is the same two solutions.

8. Originally Posted by Joker37
How did you get from this....
Joker, check this out. It will help you understand the quadratic formula in it's entirety. Remember, as e-pi has asserted, that generally all constants and variables are considered as coefficient to the quadratic term in question. EG

$2y^3zx^2+yz^4x+19yz=0$ is quadratic in x (you just have to see it). To isolate x, use $a=2y^3z,b=yz^4,\text{ and }c=19yz$.

http://www.mathhelpforum.com/math-he...c-formula.html