Quadratic Equations
We are familiar with the quadratic equation $ax^{2}+bx+c=0$, where $a\neq 0,b,c$ are real numbers and solved them in the system of real numbers in the cases where $b^{2}-4ac\geq 0$.
Now we shall solve the quadratic equation $ax^{2}+bx+c=0$, where $a\neq 0,b,c$ are real numbers and $b^{2}-4ac< 0$.
The solution to the above equation are available in the set of complex numbers which are given by,
$\large x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$ $\large =\frac{-b\pm i\sqrt{4ac-b^{2}}}{2a}$
Fundamental theorem of algebra
"A polynomial equation has at least one root".
"A polynomial equation of degree n has exactly n roots".
We are familiar with the quadratic equation $ax^{2}+bx+c=0$, where $a\neq 0,b,c$ are real numbers and solved them in the system of real numbers in the cases where $b^{2}-4ac\geq 0$.
Now we shall solve the quadratic equation $ax^{2}+bx+c=0$, where $a\neq 0,b,c$ are real numbers and $b^{2}-4ac< 0$.
The solution to the above equation are available in the set of complex numbers which are given by,
$\large x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$ $\large =\frac{-b\pm i\sqrt{4ac-b^{2}}}{2a}$
Fundamental theorem of algebra
"A polynomial equation has at least one root".
"A polynomial equation of degree n has exactly n roots".
0 Comments