Question

2x^2-6+3=0 find nature of roots and if the real roots exits find them​

Answer 1

Answer:

Consider the given equation.

2x

2

−6x+3=0

D=(−6)

2

−4×2×3

D=36−24=12>0

So, the roots are real and unequal.

Hence, this is the answer.

Answer 2

Step-by-step explanation:

Given:-

Given equation is 2x^2-6x+3=0

To find:-

Find nature of roots and if the real roots exits find tthem?

Solution:-

Given quadratic equation is 2x^2-6x+3=0

On comparing with the standard quadratic equation ax^2 +bx + c = 0

a = 2

b= -6

c = 3

We know that

To find the nature of the roots we have to find the value of the discriminant.

The discriminant of the quadratic equation ax^2 +bx + c = 0 is D= b^2-4ac

=>(-6)^2 - 4 (2)(3)

=>36 -24

=>12 > 0

Since , The discriminant is greater than zero then The roots exists and it has distinct (different) and real roots.

Given equation is 2x^2-6x+3=0

On dividing by 2 both sides

=>(2x^2/2)-(6x/2) +(3/2) = 0/2

=>x^2 -3x +(3/2) = 0

=>x^2 -3x = -3/2

=>x^2 - 2(3x)/2 = -3/2

=>x^2 -2(x)(3/2) = -3/2

On adding (3/2)^2 both sides then

=>x^2 -2(x)(3/2) +(3/2)^2 = (-3/2)+(3/2)^2

=>[x-(3/2)]^2 = (-3/2)+(9/4)

=>[x-(3/2]^2 = (-6+9)/4

=>[x-(3/2)]^2 = 3/4

=>[x-(3/2)] = ±√(3/4)

=>x-(3/2) = ±(√3) /2

=>x = (3/2)±√(3)/2

=>x = (3±√3)/2

The values of x are (3+√3)/2 and (3-√3)/2

Answer:-

Nature of the roots :-

The quadratic equation has distinct and real roots.

Roots:-

The roots of the given quadratic equation are (3+√3)/2 and (3-√3)/2

Used formulae:-

• The discriminant of the given equation tells the nature of the roots.

• Discriminant of the equation ax^2+bx+c = 0 is D = b^2-4ac.

• If D>0 ,then the roots are distinct and real .

• If D = 0 then the roots are real and equal.

• If D < 0 then the roots are no real i.e. imaginary.

• If D≥ 0 then the roots are exist for the given equation.

Used method:-

• To find the roots of the given equation by using Completing the square method.