Multiple choice

The given quadratic equations have real roots and roots are Real and equal, $\sqrt{\dfrac{3}{2}}$ : $2x^2 \, - \, 2\sqrt{6x} \, + \, 3 \, = \, 0$

  1. True

  2. False
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A Correct answer
Explanation

The discriminant of the quadratic equation 2x^2 - 2*sqrt(6)x + 3 = 0 is D = (-2*sqrt(6))^2 - 4(2)(3) = 24 - 24 = 0. Since the discriminant is 0, the roots are real and equal.

AI explanation

To determine the nature of the roots, we calculate the discriminant using the formula b squared minus 4ac for the equation 2x^2 minus 2 times the square root of 6 times x plus 3 equals 0. The discriminant evaluates to 24 minus 24, which equals 0, proving the roots are real and equal. We find the root by using the formula x equals negative b divided by 2a, giving 2 times the square root of 6 divided by 4. This simplifies to the square root of 6 divided by 2, which is equivalent to the square root of 3 divided by the square root of 2. Therefore, the statement is True.

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