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

EXPLANATION

Equation given

3x^2 - 2x + 1/3 = 0

equation will be written as:-

9x^2 - 6x + 1 = 0

TO FIND DISCRIMINANT.

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

D = (-6)^2 - 4 (9)(1) = 0

D = 36 - 36 = 0

D = 0

Roots are real

TO FIND NATURE OF ROOTS

IF THEY ARE REAL

Nature of roots are real and equal

Therefore,

D = 0

SOME RELATED FORMULA

$1) = d > 0 \: \: roots \: \: are \: \: real \: \: and \: \: unequal$

$2) = d \: = 0 \: \: roots \: \: are \: \: real \: \: and \: \: equal$

$3) = d < 0 \: \: roots \: \: are \: \: imaginary$

Answer 2

Question:-

Find the discriminate of the equation

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

and hence find the nature of its roots. Find them,if they are real.

Answer:-

Here

a = 3

b = -2

$b = \frac{1}{3}$

Therefore, discriminate

$= > {b}^{2} - 4ac$

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

$= 4 - 4$

$= 0$

Hence, the given quadratic equation has two equal roots.

The roots are

$= > \frac{ - b}{2a}$

i.e.,

$\frac{2}{6} . \frac{2}{6}$

i.e.,

$\frac{1}{3} . \: \frac{1}{3}$

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