Question

Find the discriminant of the equation and the nature of roots. Also find the roots, if they are real:

$3x^2-2x+\frac{1}{3}=0$

• A. Roots are imaginary
  
• B. D $=$ 0, Roots are real and equal $\frac{1}{3},\frac{1}{3}$
• C. D$=$ $\frac{2}{5}$, Roots are real and unequal $\frac{1}{5},\frac{1}{2}$
• D. Cannot be determined
  

Answer 1

Answer: (B)

D $=$ 0, Roots are real and equal $\frac{1}{3},\frac{1}{3}$

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

Comparing $3x^2-2x+\frac{1}{3}=0$ with $a{x}^2+bx+c=0$ we get $a=3,b=-2,c=\frac{1}{3}$

Therefore $D=b^2-4ac$
$=(-2{)}^2-4\times 3\times \frac{1}{3}$
$=4-4$
$=0$

Therefore roots are real and equal.

Therefore roots are,

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

$=\frac{-(-2)\pm \sqrt{(-2{)}^2-4\times 3\times \frac{1}{3}}}{2\times 3}$

$=\frac{2\pm 0}{2\times 3}$

$=\frac{2}{2\times 3}$

$=\frac{1}{3}$

Therefore roots are $\frac{1}{3}$ ,$\frac{1}{3}$