Question

Find the roots of the quadratic equation by splitting the middle term of √3x^2-2x-√3​

Answer 1

Answer:

$\sqrt{3} {x}^{2} - 2x - \sqrt{3} = 0 \\ \\ \sqrt{3} {x}^{2} - 2x - \sqrt{3} - ( - \sqrt{3} ) = - ( - \sqrt{3} ) \\ \\ \sqrt{3} {x}^{2} - 2x = - ( - \sqrt{3} ) \\ \\ \sqrt{3} {x}^{2} - 2x = \sqrt{3} \\ \\ \frac{ \sqrt{3} {x}^{2} - 2x }{ \sqrt{3} } = \frac{ \sqrt{3} }{ \sqrt{3} } \\ \\ {x}^{2} + ( - \frac{2 \sqrt{3} }{3} x) = 1 \\ \\ {x}^{2} + ( - \frac{2 \sqrt{3} }{3} x) + {( \frac{ - \sqrt{3} }{3} )}^{2} = 1 + {( \frac{ - \sqrt{3} }{3}) }^{2} \\ \\ {x}^{2} + ( - \frac{2 \sqrt{3} }{ 3} x) + \frac{1}{3} = \frac{4}{3} \\ \\ \sqrt{ {(x - \frac{ \sqrt{3} }{3} )}^{2} } = \sqrt{ \frac{4}{3} } \\ \\ (x - \frac{ \sqrt{3} }{3} ) = \frac{2 \sqrt{3} }{3} \: , \: (x - \frac{ \sqrt{3} }{3} ) = - \frac{2 \sqrt{3} }{3} \\ \\ x = \sqrt{3} \: , \: x = - \frac{1}{ \sqrt{3} }$

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