In Algebra, we will find certain quadratic equations with negative roots in their solutions. These kinds of roots have imaginary numbers and the roots are sometimes called the complex roots. In this article, we will discuss the complex numbers and quadratic equations and the nature of roots in detail. Show Complex Numbers and Quadratic EquationsA complex number can be represented in the form of a+bi, which is the combination of both the real numbers and the imaginary numbers. Here, a and b are real numbers and i is the imaginary number. A quadratic equation is an equation, where atleast one term should be squared. The maximum degree of the equation must be two. For example, 5x2+3x+3 =0. In this case, the highest order of the equation is 2. So, the given equation is a quadratic equation. Standard FormThe standard form of the quadratic equation is given by \(\begin{array}{l}ax^{2}+bx+c\end{array} \) Where a,b,c are real numbers and \(\begin{array}{l}a\neq 0\end{array} \) .The roots of the equation is given by- \(\begin{array}{l}x= \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\end{array} \) Discriminant (D) \(\begin{array}{l}= \sqrt{b^{2}-4ac}\end{array} \) Nature of the RootsThe root can be of three form depending upon the value of D.
given by, \(\begin{array}{l}x_{1}= \frac{-b + \sqrt{b^{2}-4ac}}{2a},x_{2}= \frac{-b – \sqrt{b^{2}-4ac}}{2a}\end{array} \)
given by, \(\begin{array}{l}x_{1},x_{2}= \frac{-b}{2a}\end{array} \)
given by, \(\begin{array}{l}x_{1}= \frac{-b + i\sqrt{4ac – b^{2}}}{2a},x_{2}= \frac{-b – i\sqrt{4ac – b^{2}}}{2a}\end{array} \) Note- A polynomial of degree n will have n roots. This is known as fundamental theorem of algebra. The quadratic equation has a degree of 2, thus they have 2 roots. Power of i\(\begin{array}{l}i^{2}= -1\end{array} \) \(\begin{array}{l}i^{3}= i.i^{2}= i(-1)=-i\end{array} \) \(\begin{array}{l}i^{4}= i^{2}.i^{2}= (-1)(-1)=1\end{array} \) Thus the general form is given by- \(\begin{array}{l}i^{4k}= i\end{array} \) \(\begin{array}{l}i^{4k+1}= -1\end{array} \) \(\begin{array}{l}i^{4k+2}= -i\end{array} \) \(\begin{array}{l}i^{4k+3}= 1\end{array} \) For Complex CaseAs we know, the discriminant of the quadratic equation is given by; D = √(b2-4ac) If 4ac > b2 ,then the solution will have non-zero imaginary part. Therefore, we can write; \(\begin{array}{l}\sqrt{\left(-\left(4 a c-b^{2}\right)\right)}=\sqrt{\left(4 a c-b^{2}\right)} \cdot \mathrm{i}\end{array} \) Now, the square root \(\begin{array}{l}\sqrt{\left(4 a c-b^{2}\right)}\end{array} \) is positive and real. Therefore, for complex number case, we get the solution for quadratic equation as;\(\begin{array}{l}\mathrm{x}=\frac{-b \pm i \cdot\left(4 a c-b^{2}\right)^{1 / 2}}{2 a}\end{array} \) Solved ExamplesExample 1: Find the roots of the quadratic equation \(\begin{array}{l}x^{2}-x+1=0\end{array} \) .Solution: Comparing the given equation with the general form of the equation We have, a= 1, b= -1, c= 1 D= \(\begin{array}{l}b^{2}-4ac = (-1)^{2}-4(1)(1)=-3\end{array} \) Thus the equation have two complex roots. \(\begin{array}{l}x= \frac{-b\pm \sqrt{D}}{2a}\end{array} \) \(\begin{array}{l}x=\frac{-1\pm \sqrt{-3}}{2(1)}\end{array} \) \(\begin{array}{l}x=\frac{-1\pm \sqrt{3}i}{2}\end{array} \) Thus the roots are \(\begin{array}{l}x=\frac{-1 + \sqrt{3}i}{2}, \frac{-1 – \sqrt{3}i}{2}\end{array} \) Example 2: Solve \(\begin{array}{l}\sqrt{5}x^{2}+x+\sqrt{5}= 0\end{array} \) Solution: Comparing the given equation with the general form of the equation We have, \(\begin{array}{l}a = \sqrt{5},b = 1, c = \sqrt{5}\end{array} \) Here Discriminant (D) = \(\begin{array}{l}b^{2} – 4ac = (1)^{2}- 4(\sqrt{5}).(\sqrt{5})= -19\end{array} \) \(\begin{array}{l}x= \frac{-b\pm \sqrt{D}}{2a}\end{array} \) \(\begin{array}{l}x= \frac{-1\pm \sqrt{-19}}{2\sqrt{5}}\end{array} \) \(\begin{array}{l}x= \frac{-1\pm \sqrt{19}i}{2\sqrt{5}}\end{array} \) Can a quadratic equation have imaginary numbers?In relation to quadratic equations, imaginary numbers (and complex roots) occur when the value under the radical portion of the quadratic formula is negative. When this occurs, the equation has no roots (or zeros) in the set of real numbers.
|