JEE Main & Advanced Mathematics Complex Numbers and Quadratic Equations Solution of Quadratic Equation

(1) Factorization method

Let \[a{{x}^{2}}+bx+c=\]\[a(x-\alpha )(x-\beta )=0\].

Then \[x=\alpha \] and \[x=\beta \] will satisfy the given equation.

Hence, factorize the equation and equating each factor to zero gives roots of the equation.

Example : \[3{{x}^{2}}-2x+1=0\] \[\Rightarrow \]\[(x-1)(3x+1)=0\];

\[x=1,\,-1/3\]

(2) Sri Dharacharya method : By completing the perfect square as  \[a{{x}^{2}}+bx+c=0\]\[\Rightarrow \]\[{{x}^{2}}+\frac{b}{a}x+\frac{c}{a}=0\]

Adding and subtracting \[{{\left( \frac{b}{2a} \right)}^{2}}\], \[\left[ {{\left( x+\frac{b}{2a} \right)}^{2}}-\frac{{{b}^{2}}-4ac}{4{{a}^{2}}} \right]=0\]   which gives, \[x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].

Hence the quadratic equation \[a{{x}^{2}}+bx+c=0\] \[(a\ne 0)\] has two roots, given by \[\alpha =\frac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}\], \[\beta =\frac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}\]

Every quadratic equation has two and only two roots.