Using the quadratic formula, find the exact value of x for the following.
-2kx^2 + 4x + 6k = 0
The quadratic formula is $\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
In your case - assuming k is a constant,
$\displaystyle
a = -2k
b = 4
c = 6k
$.
Just put these into your answer and solve for x. Your answer will have $\displaystyle k$ in it, the "exact value" will most probably involve surds.
Hope this helps
If you have an equation that says $\displaystyle ax^2+bx+c=0$ then the quadratic formula is as follows (for $\displaystyle k \neq f(x)$):
$\displaystyle x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$
- 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:
$\displaystyle x = \frac{-4 \pm \sqrt{16- (4 \times -2k \times 6k)}}{-4k}$
Simplify to give $\displaystyle 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
$\displaystyle x = \frac{-4 \pm \sqrt{16- (4 \times -2k \times 6k)}}{-4k}$
Collecting and multiplying out the terms in the square root gives $\displaystyle 16-(-48k^2) = 48k^2+16$
As the HCF of 48 and 16 is 16 we can remove this as a factor: $\displaystyle 48k^2+16 = 16(3k^2+1)$ note, I made a mistake here in my other working - that +3 should have been a +1
Reintroducing this to the full equation:
$\displaystyle \frac{-4 \pm \sqrt{16(3k^2+1)}}{-4k}$
As $\displaystyle \sqrt{ab} = \sqrt{a} \sqrt{b}$ then $\displaystyle \sqrt{16(3k^2+1)} = \sqrt{16} \times \sqrt{3k^2+1} = 4\sqrt{3k^2+1}$
Reintroducing the simplified discriminant:
$\displaystyle \frac{-4 \pm 4\sqrt{3k^2+1}}{-4k} = \frac{4(-1 \pm \sqrt{3k^2+1})}{-4k}$
4 will cancel to give
$\displaystyle \frac{-1 \pm \sqrt{3k^2+1}}{-k}$
You can leave it like that but it's more usual to not have a negative denominator so we multiply by $\displaystyle \frac{-1}{-1}$
$\displaystyle -\frac{-1 \pm \sqrt{3k^2+1}}{-k} = \frac{1 \mp \sqrt{3k^2+1}}{k}$
Since there is a kČ term under the square root sign then x will be real for all real k
The same numbers but in different order!
Notice that he had first $\displaystyle \frac{-1\pm\sqrt{3k^2+1}}{-k}$ which gives the two solutions $\displaystyle \frac{-1+\sqrt{3k^2+1}}{-k}= \frac{1-\sqrt{3k^2+1}}{k}$ and $\displaystyle \frac{-1- \sqrt{3k^2+1}}{-k}= \frac{1+\sqrt{k^2+1}}{k}$.
e^(i*pi) just changed from $\displaystyle \pm$ to $\displaystyle \mp$ to make it clear that the $\displaystyle -\sqrt{k^2+1}$ had changed to $\displaystyle +\sqrt{k^2+1}$ and vice-versa. Of course, it is the same two solutions.
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
$\displaystyle 2y^3zx^2+yz^4x+19yz=0$ is quadratic in x (you just have to see it). To isolate x, use $\displaystyle a=2y^3z,b=yz^4,\text{ and }c=19yz$.
http://www.mathhelpforum.com/math-he...c-formula.html