Factoring a quadratic equation code#
Factoring a quadratic equation plus#

In standard form like this, that the roots of it are So let's do thatįormula tells us that if we have something The square, or we could apply the quadraticįormula, which is really just a formula derivedįrom completing the square. The x term, but I would get 5/2 for the constant. Would get integer coefficients on the x squared in To factor it, I would divide both sides by 2. Two characters cancel out, and we just are left with 0. Hand side becomes 2x squared minus 6x plusĥ is equal to- and then on our right hand side, these Have to take the 6x and get rid of it from

Factoring a quadratic equation plus#

And standard form, ofĬourse, is the form ax squared plus bx plusĬ is equal to 0. That we're more familiar with, let's try to put it Long story short: the roots will NOT be the same as the first equation! Hope that helped! X = (-B +- sqrt(B^2 + 4AC))/2A (remember, minus -C^2 is the same as plus C^2)Ĭompare this to the solution of our original equation:Īs long as A, B, and C are not zero (you're dealing with an actual quadratic equation), you can see that x is different. Our equation looks the same as the original, except C is now different. Put it together and we get Ax^2 + Bx + (C/i) = 0. Now, let's try dividing the whole equation by 'i': even though it's imaginary, it's still a number, right? iAx^2 becomes Ax^2, iBx becomes Bx, C becomes (C/i), and 0 becomes 0 (0 divided by ANY number, even imaginary ones, is always equal to 0). Think about it! So if you have some quadratic equation Ax^2 + Bx + C = 0, and you multiply each 'x' term by i, you'd get iAx^2 + iBx C = 0. Note: x+2pi*k just refers to any angle coterminal with x, so they are equivalent so what about the +2pi*k? Well, substitute x with x+2pi*k and voila! (In that first case, n represents any old number, but with 1/n, n is the number of roots) Keeping that in mind, let's arbitrarily substitute n for 1/n. Technically (cos x + i*sin x)^n = (r*(cos x + i*sin x))^n, but the r is omitted because we're usually talking about the unit circle (r=1).

Clearly a very different formula, but that doesn't mean it can't be derived from that first expression. I don't know if this is how you've seen it, but the version of de Moivre's theorem in my textbook is r^(1/n)*(cos((x + 2pi*k)/n) + i*sin((x + 2pi*k)/n)). I can't seem to find any videos on de Moivre's theorem, either, but I do know that the idea that (cos x + i*sin x)^n = cos(nx) + i*sin(nx) can be derived from Euler's formula. The red line is the parabola that you normally get when using only real coordinates Plot3(realPart, zeros(1,numberOfPointsOnTheAxis). ImaginaryPart(:) * ones(1,numberOfPointsOnTheAxis) * 1i X = ones(numberOfPointsOnTheAxis,1) * realPart(:)' +. ImaginaryPart = linspace(-2,2,numberOfPointsOnTheAxis) RealPart = linspace(1,4,numberOfPointsOnTheAxis) P = % 1*x*x -6*x +10 the coefficients of the polynomial

Factoring a quadratic equation code#

I came up with the following piece of code written in MATLAB: You could make two representations, one for the real value of the result and one for the imaginary value of the result, but you would have to search for the point(s) where those 2 are both 0.

The absolute value is always non-negative, and the solutions to the polynomial are located at the points where the absolute value of the result is 0. You would put the absolute value of the result on the z-axis when x is real (complex part is 0) the absolute value is equal to the value of the polynomial at that point. I think a way to do that is to make a 3D chart that has the complex coordinates on the horizontal axis.