Question

The Discriminet of the quadratic equation 3x ^ 2 + 1/3 - 2x ​

Answer 1

Answer :

0

Solution :

Here ,

The given quadratic polynomial is ;

3x² - ⅓ - 2x

Now ,

Rearranging the terms of the given quadratic polynomial in its general form , we have ;

3x² - 2x - ⅓

Now ,

Comparing the above equation with the general quadratic equation ax² + bx + c , we have ;

a = 3

b = -2

c = ⅓

Now ,

The discriminant of the quadratic polynomial will be given as ;

=> D = b² - 4ac

=> D = (-2)² - 4•3•⅓

=> D = 4 - 4

=> D = 0

Hence ,

The discriminant of the given quadratic polynomial , D = 0 .

Moreover ,

Since discriminant D = 0 , thus the given quadratic polynomial has equal zeros .

Answer 2

Answer:

$\huge\rm{Solution:–}$

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

$\rm \: This \: equation \: is \: of \: the \: form \: ax²+bx+c=0 \\ \rm \: a=3,b=-2 \: and \: c= \frac{1}{3}$

$\rm \: Discriminant=b²-4ac \\ \rm→ b²-4ac=(-2)-4(3)( \frac{1}{3} )$

$\rm \: or \: {b}^{2} - 4ac = 4 - 4 \\ \rm \: or \: {b}^{2} - 4ac = 0$

$\rm \: If \: b²-4ac>0→Then \: we \: will \: get \: two \: different \:real \: roots$

$\rm \: If \: b²-4ac=0→ we \: get \: two \: same \: roots \\ \rm \: If \: b²-4ac<0→ We \: will \: not \: get \: real \: roots$

$\rm{\underline{\underline{Quadratic \: Formula:–}}}$

$\rm \: x= \frac{-b± \sqrt{b²-4ac} }{2a} \\ \rm \: x= \frac{-(-2)± \sqrt{0} }{2(3)}$

$\rm \: or \: x= \frac{2±0}{6} \\ \rm \: x= \frac{2-0}{6} \: and \: x= \frac{2+0}{6}$

$\rm \: or \: x= \frac{2}{6} \: and \: x= \frac{2}{6} \\ \rm \: or \: x= \frac{1}{3} \: and \: x= \frac{1}{3}$