Question

Solve the value of x: root3x^2-2root2x-2root3=0

Answer 1

Answer:

i hope you are satisfied with my solution

Answer 2

$\red\bigstar$CORRECT QUESTION :-

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Solve the value of x:

$\sf \sqrt{3} {x}^{2} - 2 \sqrt{2} x - 2 \sqrt{2} = 0$

$\blue\bigstar$FORMULA REQUIRED:-

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•For finding Discriminant

$\boxed{ \bf \circ \: D = b^{2} - 4ac}$

•For finding value of x

$\boxed{ \bf \circ \: x = \dfrac{ - b \pm \sqrt{D} }{2a} }$

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$\pink\bigstar$SOLUTION:-

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$\sf Given:- \sf \sqrt{3} {x}^{2} - 2 \sqrt{2} x - 2 \sqrt{2} = 0$

It is in the form of ax²+bx+c=0

$\sf where,a= \sqrt{3} ,b= -2\sqrt{2} \: and \: c= - 2 \sqrt{3}$

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We know that

$\sf D = b^{2} - 4ac$

$\sf D =( - 2 \sqrt{2} )^{2} - 4( \sqrt{3} )( - 2 \sqrt{3} )$

$\sf \implies8 + 24$

$\implies \sf32 > 0$

$\sf \therefore x = \dfrac{ - b \pm \sqrt{D} }{2a}$

$\sf x = \dfrac{ - (2 \sqrt{2 }) \pm \sqrt{32} }{2 \times \sqrt{3} } \\ \\ = \sf \dfrac{2 \sqrt{2} \pm 4 \sqrt{2} }{2 \sqrt{3} } \: \: \:$

$\implies \sf x = \dfrac{2 \sqrt{2} + 4 \sqrt{2} }{2 \sqrt{3} } \: or \: x = \dfrac{2 \sqrt{2} - 4 \sqrt{2} }{2 \sqrt{3} }$

$\implies \sf x = \dfrac{6 \sqrt{2} }{2 \sqrt{3} } \: or \: x = \dfrac{ - 2 \sqrt{2} }{2 \sqrt{3} }$

•Rationalizing the denominators

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$\implies \sf x = \dfrac{3 \sqrt{2} }{3} \times \dfrac{ \sqrt{ 3} }{ \sqrt{3} } \: or \: x = \dfrac{ - \sqrt{2} }{ \sqrt{3} } \times \dfrac{ \sqrt{3} }{ \sqrt{3} }$

$\implies \sf x = \sqrt{6} \: or \: x = \dfrac{ - 1}{3} \sqrt{6}$

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Hence,the roots of the given equations are:-

$\bf \sqrt{6} \: and \: \dfrac{ - 1}{3} \sqrt{6}$