What is the nature of the roots of √3 x^2+6x-√3=0 ?

Correct option is D

Given:

​$$\sqrt{3}x^2 + 6x - \sqrt{3} = 0$$

Solution:

To calculate the nature of the root of the given equation, we calculate the discriminant (D)

The discriminant of the quadradic equation in the form of $$ax^2 + bx + c = 0$$ is :

Discriminant(D) = $$b^2-4ac$$ 

Here, $$a= \sqrt3 , b= 6, c = -\sqrt3$$ 

Now put in the value of $$a,b\ and\ c$$, we get

D= $$6^2 - 4(\sqrt{3})(-\sqrt{3})$$

$$= 36 - 4(-3)$$

$$= 36 + 12 = 48$$

Since D$$> 0, \text{ the roots are real and distinct.}$$

Thus the correct answer is (D)

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