Quadratic equation in complex numbers
A complex number can be represented in the form of a+bi,
where a and b are real numbers and i is the imaginary number.
Quadratic Equation-
Standard Form-
The standard form of quadratic equation is given by
| \(ax^{2}+bx+c\) |
Where a,b,c are real numbers and \(a\neq 0\).
The roots of the equation is given by-
| \(x= \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\) |
Discriminant (D) \(= \sqrt{b^{2}-4ac}\)
The root can be of three form depending upon the value of D.
- Distinct Roots- When D>0,then the quadratic equation has two distinct roots.
given by, \(x_{1}= \frac{-b + \sqrt{b^{2}-4ac}}{2a},x_{2}= \frac{-b – \sqrt{b^{2}-4ac}}{2a}\)
- Equal Roots- When D=0,then the quadratic equation has two equal roots.
given by, \(x_{1},x_{2}= \frac{-b}{2a}\)
- Complex Roots- When D<0, then the quadratic equation has two complex roots.
given by, \(x_{1}= \frac{-b + i\sqrt{4ac – b^{2}}}{2a},x_{2}= \frac{-b – i\sqrt{4ac – b^{2}}}{2a}\)
Note- A polynomial of degree n will have n roots. This is known as fundamental theorem of algebra.
The quadratic equation has degree of 2, thus they have 2 roots.
Power of i-
\(i^{2}= -1\)
\(i^{3}= i.i^{2}= i(-1)=-i\)
\(i^{4}= i^{2}.i^{2}= (-1)(-1)=1\)
Thus the general form is given by-
\(i^{4k}= i\)
\(i^{4k+1}= -1\)
\(i^{4k+2}= -i\)
\(i^{4k+3}= 1\)
Example- Find the roots of the quadratic equation \(x^{2}-x+1=0\).
Solution- Comparing the given equation with the general form of the equation
We have, a= 1, b= -1, c= 1
D= \(b^{2}-4ac = (-1)^{2}-4(1)(1)=-3\)
Thus the equation have two complex roots.
\(x= \frac{-b\pm \sqrt{D}}{2a}\)
\(x=\frac{-1\pm \sqrt{-3}}{2(1)}\)
\(x=\frac{-1\pm \sqrt{3}i}{2}\)
Thus the roots are \(x=\frac{-1 + \sqrt{3}i}{2}, \frac{-1 – \sqrt{3}i}{2}\)
Example- Solve \(\sqrt{5}x^{2}+x+\sqrt{5}= 0\)
Solution- Comparing the given equation with the general form of the equation
We have, \(a = \sqrt{5},b = 1, c = \sqrt{5}\)
Here Discriminant (D) = \(b^{2} – 4ac = (1)^{2}- 4(\sqrt{5}).(\sqrt{5})= -19\)
\(x= \frac{-b\pm \sqrt{D}}{2a}\)
\(x= \frac{-1\pm \sqrt{-19}}{2\sqrt{5}}\)
\(x= \frac{-1\pm \sqrt{19}i}{2\sqrt{5}}\)<
To learn more about NCERT Solutions for Complex Numbers and Quadratic equations click on this link, NCERT Solutions for Complex Numbers and Quadratic equations.