Question

$\sqrt{3x { } } - 8x + 4 \sqrt{3}$
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Answer 1

Correct Question:-

$\sqrt{3 {x}^{2} } - 8x + 4 \sqrt{3}$

Step By Step Solution:-

√3X² - 8X + √3 = 0

AX² + BX + C = 0

A = √3, B = -8, C = 4√3

X = -B + √B² -4 AC

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..............2A...................

$= ( -8 ) + \sqrt{( -8 )² -4 × \sqrt{3 \times 4 \sqrt{3} } }$

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................................2 × √4...................................

$= 8 + \frac{\sqrt{64 - 16 \times 3} }{2 \sqrt{3} }$

$= 8 + \frac{\sqrt{64 - 48} }{2 \sqrt{3} }$

$\frac{ = 8 + 16}{2 \sqrt{3} } = \frac{8 + 4}{2 \sqrt{3} }$

$= \frac{2(4 + 2)}{2 \sqrt{3} }$

$x = \frac{4 + 2}{ \sqrt{3} }$

$x = \frac{4 + 2}{ \sqrt{3} } . \: x = \frac{4 - 2}{ \sqrt{3} }$

$x = \frac{6}{ \sqrt{3} } . \: x = \frac{2}{ \sqrt{3} }$

Done.

Answer 2

Answer:

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

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