Quadratic Formula

Back to the main page

All quadratic equations are represented in the following format: $a{x}^2+bx+c=0$
And if there are terms on the right-hand side, move them to the left-hand side like this: $$a{x}^2=bx+c$$ $$a{x}^2-bx-c=0$$ The signs of the coefficients a, b and the constant c may be any combinations of negative or positive signs.

No we will solve the following equation: $\mathrm{16}\cdot x^2+5\cdot x-7=0$
Always start by dividing both sides of the equals sign by a. In this case a is 16. This yields: $$\frac{\mathrm{16}{x}^2}{\mathrm{16}}+\frac{5x}{\mathrm{16}}-\frac{7}{\mathrm{16}}=0$$ Or in other terms: $$x^2+\frac{5x}{\mathrm{16}}-\frac{7}{\mathrm{16}}=0$$ That makes the coefficient of the $x^2$-term 1, which is what we want.

The quadratic formula reads as follows:
Given a quadratic equation of the form $$a{x}^2+bx+c=0$$ $$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
Note the $\pm$ sign. It has two solutions. One is called $x_1$. The other one is called $x_2$. You get $x_1$ by replacing the $\pm$ sign by a plus sign. You get $x_2$ by replacing the $\pm$ sign by a minus sign.

The quadratic formula can also be written it like this:
Given a quadratic equation of the form $$x^2+px+q=0$$ $$x=-\frac{p}{2}\pm \sqrt{\frac{p^2}{4}-q}$$ In our case $p=\frac{5}{\mathrm{16}}$ and $q=-\frac{7}{\mathrm{16}}$

Using the second version of the quadratic formula we get: $$x=-\frac{\frac{5}{\mathrm{16}}}{2}\pm \sqrt{\frac{{\left(\frac{5}{\mathrm{16}}\right)}^2}{4}+\frac{7}{\mathrm{16}}}$$ which is equal to $$x=-\frac{5}{\mathrm{32}}\pm \sqrt{\frac{\frac{\mathrm{25}}{{\mathrm{16}}^2}}{4}+\frac{7}{\mathrm{16}}}$$ which is equal to $$x=-\frac{5}{8\cdot 4}\pm \sqrt{\frac{\frac{\mathrm{25}}{4^4}}{4}+\frac{7}{4^2}}$$ which is equal to $$x=-\frac{5}{8\cdot 4}\pm \sqrt{\frac{\frac{\mathrm{25}}{4^3}+7}{4^2}}$$ which is equal to $$x=\frac{-\frac{5}{8}\pm \sqrt{\frac{\mathrm{25}}{\mathrm{64}}+7}}{4}$$
So $$x_1\approx \mathrm{0.523}$$ and $$x_2\approx \mathrm{-0.836}$$ And that's how uneven these solutions usually are.