Question

Find the discriminant of the quadratic equation 3x2 - 2x + 1/3= 0 and hence find the
nature of its roots, if they are real.​

Answer 1

Solution :

By using Discriminant formula

★ d = b² - 4ac

⇒ d = ( - 2 )² - 4 × 3 × 1/3

⇒ d = 4 - 4

⇒ d = 0

Here d = 0 , So given equation has 1 real root

Now by using quadratic formula

★ x = ( - b ± √d ) / 2a

⇒ x = [ - ( - 2 ) ± √0 ] / 2 × 3

⇒ x = ( 2 ± 0 ) / 6

Taking +ve sign

⇒ x = ( 2 + 0 ) / 6

⇒ x = 1 / 3

Taking -ve sign

⇒ x = ( 2 - 0 ) / 6

⇒ x = 1 / 3

Answer 2

Step-by-step explanation:

$\huge{\underline{\red{AnswEr}}}$

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

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

$\implies {9x}^{2} - 6x + 1 = 0$

Comparing equations with ax²+bx+c=9

a=9

b=-6

c=1

we know that,

D=b²-4ac

D=(-6)²-4×9×1

D=36-36

D=0

since D=0

The given equations has two equal real roots.

Now,

using quadratic formula to find root

x=-b±√D/2a

x=-(-6)±√0/18

x=6/18

x=1/3

$\therefore \: the \: roots \: of \: equations \: are \: \frac{1}{3} \: and \: \frac{1}{3}$