Question

$\sqrt{3} {m }^{2} - 8m + \sqrt{48 } = 0$
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Answer 1

Answer:

$m=2\sqrt{3} ,m= \frac{2\sqrt{3} }{3}$

Step-by-step explanation:

$\sqrt{3}m^2 - 8m + \sqrt{48} = 0$

now, let us use quadratic formula,

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

discriminant:

                              $\sqrt{b^2-4ac} = \sqrt{(-8)^2-4\sqrt{3}\cdot4\sqrt3 } = \sqrt{16} =4$

now, $4 >0$ so the equation has two distinct real roots.

now using quadratic formula:

$m = \frac{8+4}{2\sqrt{3} } =\frac{12}{2\sqrt{3} }=\frac{12\sqrt{3} }{2\sqrt{3} \cdot\sqrt{3} } = 2\sqrt{3}$

or

$m = \frac{8-4}{2\sqrt{3} } = \frac{4}{2\sqrt{3} } = \frac{2}{\sqrt{3} } =\frac{2\sqrt{3} }{\sqrt{3} \cdot\sqrt{3} } = \frac{2\sqrt{3} }{3}$

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