Quadratic

Welcome to the hidden page that is not listed on the home page or the dropdown menu. Anyways, let's do the

quadratic

\[ ax^2 + bx + c = (d) x^2 + (-d(s_1 + s_2)) x + (d s_1 s_2) = d(x - s_1)(x - s_2) \]

now, I will say that if these are equal, than they have the same terms, so I will use the first to simplify, the second to solve for \(i\), and the third to solve for \(k^2\) (I'll explain later)

\[ a = d \]

\[ x^2 + \frac{b}{a} x + \frac{c}{a} = (x - s_1)(x - s_2) = x^2 + (-(s_1 + s_2)) x + (s_1 s_2) \]

\[ s_1 = i + j \]

\[ s_2 = i - j \]

\[ as_1^2 + bs_1 + c = a(s_1 - s_1)(s_1 - s_2) = a0(s_1 - s_2) = 0 \]

\[ as_2^2 + bs_2 + c = a(s_2 - s_1)(s_2 - s_2) = a(s_2 - s_1)0 = 0 \]

\[ as_1^2 + bs_1 + c = 0 \]

\[ as_2^2 + bs_2 + c = 0 \]

\[ as^2 + bs + c = 0 \]

\[ s = i \pm j = i \pm \sqrt{j^2} = i + k \]

\[ k^2 = j^2 \]

\[ \frac{b}{a} = -(s_1 + s_2) \]

\[ \frac{b}{2a} = -\frac{s_1 + s_2}{2} \]

\[ -\frac{b}{2a} = \frac{s_1 + s_2}{2} = i \]

\[ i = -\frac{b}{2a} \]

\[ \frac{c}{a} = s_1 s_2 = (i + j)(i - j) = i^2 - j^2 = \frac{b^2}{4a^2} - j^2 \]

\[ j^2 = \frac{b^2}{4a^2} - \frac{c}{a} = \frac{b^2 - 4ac}{4a^2} \]

\[ s = k - \frac{b}{2a} \]

\[ k^2 = \frac{b^2 - 4ac}{4a^2} \]

\[ s = -\frac{b}{2a} \pm \sqrt{\frac{b^2 - 4ac}{4a^2}} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

And time to continue like a multiple choice book. If you want to see the other more common proof of the quadratic formula, click here. But if you think you had enough quadratic for the day, click here