Find the roots of the following quadratic equ | vedclass

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(C) Comparing the given equation $2x^{2} - 2\sqrt{6}x + 3 = 0$ with the standard form $ax^{2} + bx + c = 0$,we get $a = 2$,$b = -2\sqrt{6}$,and $c = 3

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$\frac{\sqrt{6}}{2}$ and $\frac{\sqrt{6}}{2}$

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(C) Comparing the given equation $2x^{2} - 2\sqrt{6}x + 3 = 0$ with the standard form $ax^{2} + bx + c = 0$,we get $a = 2$,$b = -2\sqrt{6}$,and $c = 3$.First,calculate the discriminant $D = b^{2} - 4ac$:$D = (-2\sqrt{6})^{2} - 4(2)(3)$$D = 24 - 24 = 0$Since $D = 0$,the quadratic equation has two equal real roots.The roots are given by the formula $x = \frac{-b \pm \sqrt{D}}{2a}$.Since $D = 0$,$x = \frac{-b}{2a}$.$x = \frac{-(-2\sqrt{6})}{2(2)} = \frac{2\sqrt{6}}{4} = \frac{\sqrt{6}}{2}$.Thus,the roots are $\frac{\sqrt{6}}{2}$ and $\frac{\sqrt{6}}{2}$.

Find the roots of the following quadratic equation by using the quadratic formula,if they exist: $2x^{2} - 2\sqrt{6}x + 3 = 0$

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