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NFMZombie

19.11.2020

Mathematics

Quadratic formula gives the solutions to the equation $ax^2+bx+c = 0,\ a\neq 0$ . the solutions are $x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ here is an example: $2x^2-4x = 11$ we cannot use the quadratic formula until it is in $ax^2+bx+c = 0$ form. move 11 so that the right side can be 0. subtract 11 from both sides $\begin{aligned} 2x^2-4x & = 11 \\ 2x^2 - 4x - 11 & = 0 \\ (2)x^2 + (-4)x + (-11) & = 0 \end{aligned}$ to use this in the quadratic formula, we use $\begin{aligned} a & = 2 \\ b & = -4 \\ c & = -11 \end{aligned}$ therefore, plug those parameters in and then tidy things up by multiplying, etc. $\begin{aligned} x & = \frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ & = \frac{)\pm\sqrt{(-4)^2-4()} }{2(2)} \\ & = \frac{4\pm\sqrt{)} }{4} \\ & = \frac{4\pm\sqrt{16+88} }{4} \\ & = \frac{4\pm\sqrt{104} }{4} \end{aligned}$ we are able to simplify √104 $\sqrt{104} = \sqrt{4 \cdot 26} = \sqrt{4} \cdot \sqrt{26} = 2\sqrt{26}$ using this simplification: $\begin{aligned} x & = \frac{4 \pm 2\sqrt{26}}{4} \\ & = \frac{2\left(2 \pm \sqrt{26}\right)}{4} \\ & = \frac{2 \pm \sqrt{26}}{2} \end{aligned}$ so the solutions are $x = \dfrac{2 + \sqrt{26}}{2},\ \dfrac{2 - \sqrt{26}}{2}$

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