Question

Find the discriminant of the equation and the nature of roots. Also find the roots.

$2x^2-6x+3=0$

• A. $D=12$, Roots are real and distinct, $x=\frac{3+\sqrt{3}}{2},\frac{3-\sqrt{3}}{2}$
• B. $D=0$, Roots are real and equal, $x=\frac{3+\sqrt{3}}{2}$
• C. $D=-9$, Roots are imaginary.
  
• D. None of these
  

Answer 1

Answer: (A)

$D=12$, Roots are real and distinct, $x=\frac{3+\sqrt{3}}{2},\frac{3-\sqrt{3}}{2}$

Nature of the roots of a quadratic equation is determined by its discriminant $D=b^2-4ac$

Comparing $2x^2+5\sqrt{3}x+6=0$ with $a{x}^2+bx+c=0$, we get $a=2,b=-6,c=3$

Therefore, $D=b^2-4ac$
$=(-6{)}^2-4\times 2\times 3$
$=36-24$
$=12$ $>0$

Therefore roots are real.

Therefore roots are,

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$=\frac{-6\pm \sqrt{(-6{)}^2-4\times 2\times 3}}{2\times 2}$

$=\frac{6\pm \sqrt{36-24}}{4}$

$=\frac{6\pm \sqrt{12}}{4}$

$=\frac{6\pm \sqrt{2\times 2\times 3}}{4}$

$=\frac{6\pm 2\sqrt{3}}{4}$

$=\frac{2(3+\sqrt{3})}{4}$ and $=\frac{2(3-\sqrt{3})}{4}$

$=\frac{(3+\sqrt{3})}{2}$ and $=\frac{(3-\sqrt{3})}{2}$