 Hi, welcome to the session. I am Deepika here. Let's discuss the question. Question says, find the nature of the roots of the following quadratic equation. If the real roots exist, find them. Quadraty equation is 3x square minus 4 root 3x plus 4 is equal to 0. We know that the roots of the equation Ax square plus Bx plus C is equal to 0 or given by x is equal to minus B plus minus B square minus 4 is equal to 2A. If B square minus 4 is greater than 0, two distinct real roots we get. If B square minus 4 is equal to 0, two equal real roots we get. And if B square minus 4 is less than 0, no real roots exist. So let's start the solution. Our given quadratic equation is 3x square minus 4 root 3x plus 4 is equal to 0. Compare this quadratic equation with Ax square plus Bx plus C is equal to 0. We get A is equal to 3, B is equal to minus 4 root 3 and C is equal to 4. Therefore the discriminant is that is B square minus 4ac is equal to minus 4 root 3 whole square minus 4 into A that is 3 into 4. Which is equal to 48 minus 48. Hence our discriminant is equal to 0 because B square minus 4ac is equal to 0. Therefore two equal real roots will exist. Now we will find the roots of this quadratic equation. Let alpha and beta are the roots of the given quadratic equation. Therefore alpha is equal to minus B plus discriminant that is B square minus 4ac upon 2A. And beta is equal to minus B minus B square minus 4ac upon 2A. Therefore alpha is equal to minus of minus 4 root 3 plus 0 upon 2 into 3. And beta is equal to minus of minus 4 root 3 minus 0 upon 2 into 3. Therefore alpha is equal to 4 root 3 upon 6 and beta is equal to 4 root 3 upon 6. Again on cancellation we have alpha is equal to 2 root 3 upon 3 and beta is equal to 2 root 3 upon 3. Which is again equal to 2 upon root 3 and beta is equal to 2 upon root 3. Hence for the above quadratic equation we have equal roots which are 2 by root 3 and 2 by root 3. This is our answer. I hope you understood the question. Bye and have a good day.