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  • Solve this problem - Complex numbers and quadratic equations - JEE Main-2

Roots of equation 2x^{2}-6x+3= 0 are 

  • Option 1)

    \frac{2\pm \sqrt{3}}{2}

  • Option 2)

    \frac{3\pm \sqrt{2}}{2}

  • Option 3)

    \frac{3\pm \sqrt{3}}{2}

  • Option 4)

    \frac{-3\pm \sqrt{3}}{2}

 

Answers (1)

best_answer

\because \alpha ,\beta =\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}

\because \alpha ,\beta =\frac{6\pm \sqrt{36-4(2)(3)}}{4}

\because \alpha ,\beta = \frac{6\pm \sqrt{12}}{4}

= \because \alpha ,\beta = \frac{3\pm \sqrt{3}}{2}

 

Roots of Quadratic Equation -

\alpha = \frac{-b+\sqrt{b^{2}-4ac}}{2a}

\beta = \frac{-b-\sqrt{b^{2}-4ac}}{2a}
 

 

- wherein

ax^{2}+bx+c= 0

is the equation

a,b,c\in R,\: \: a\neq 0

 

 


Option 1)

\frac{2\pm \sqrt{3}}{2}

This is incorrect

Option 2)

\frac{3\pm \sqrt{2}}{2}

This is incorrect

Option 3)

\frac{3\pm \sqrt{3}}{2}

This is correct

Option 4)

\frac{-3\pm \sqrt{3}}{2}

This is incorrect

Posted by

Himanshu

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