- Joined
- Nov 29, 2014
I can't even believe I'm actually doing this. @Autphag, this is pretty basic, so try to follow up.
First, with trained eye we can see that what @Marshal Mannerheim said is true. We have simple identity, which follows:
We can prove this by taking those two terms, and multiplying them with each others.
Hence, if we use that on another direction, we'll get the reduction @Marshall Mannerheim proposed.
Here's also another way to solve that quadratic equation, using yet another formula. (if you'd excuse my writing). We could also prove this formula, but since I don't really have time or interest (and I lack blackboard), so I'll just refer to wikipedia.
Note that there's often two instances where quadratic equation is zero. This is easiest to demonstrate by drawing, since the graph for quadratic equations is parabolic. And we can see
that if there's such a real number that f(x)=0, it can be two of those.
And here's the actual solution, by using the said formula.
Please note that discriminant, aka the b^2 - 4ac part here is zero. This means we'll have dual root, which means that there's only one real value for x where the f(x) = 0. I've also drawn a simple graph to visualize this.
Please, feel free to ask me if you have difficulties to follow this. Class dismissed.
Code:
(x-a)(x-b) = x^2 - (a+b)x + ab
Code:
(x-a)(x-b)
=x(x-b)-a(x-b)
=x^2 + x*-b - a*x + ab
=x^2 + (-a-b)*x + ab
=x^2 - (a+b)x + ab
Here's also another way to solve that quadratic equation, using yet another formula. (if you'd excuse my writing). We could also prove this formula, but since I don't really have time or interest (and I lack blackboard), so I'll just refer to wikipedia.
Note that there's often two instances where quadratic equation is zero. This is easiest to demonstrate by drawing, since the graph for quadratic equations is parabolic. And we can see
that if there's such a real number that f(x)=0, it can be two of those.
And here's the actual solution, by using the said formula.
Please note that discriminant, aka the b^2 - 4ac part here is zero. This means we'll have dual root, which means that there's only one real value for x where the f(x) = 0. I've also drawn a simple graph to visualize this.
Please, feel free to ask me if you have difficulties to follow this. Class dismissed.
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