Solution of Quadratic Equation
Category : JEE Main & Advanced
(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.
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