Math, asked by pinky2845, 1 year ago

the roots of that equation are

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Answered by hukam0685
0

 \sqrt{3}  {x}^{2}  + 10x - 8 \sqrt{3}  \\  \sqrt{3}  {x}^{2}  + 12x - 2x - 8 \sqrt{3}  \\  \sqrt{3}  {x}^{2}  + 4 \times 3x - 2x - 2 \times 4 \sqrt{3}  \\ \sqrt{3}  {x}^{2}  + 4 \times  \sqrt{3} \sqrt{3}  x - 2x - 2 \times 4 \sqrt{3} \\  \sqrt{3}x(x + 4 \sqrt{3})   - 2(x + 4 \sqrt{3} ) \\ (x + 4 \sqrt{3} )( \sqrt{3} x - 2) = 0 \\ (x + 4 \sqrt{3} ) = 0 \\ x  = - 4 \sqrt{3}  \\ ( \sqrt{3} x - 2) = 0 \\  \sqrt{3x}  = 2 \\ x =  \frac{2}{ \sqrt{3} }  \\
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