Math, asked by lavanyauday6614, 8 months ago

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

Answers

Answered by mishradeeksha273
1

Answer:

i hope you are satisfied with my solution

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Answered by Anonymous
118

\red\bigstarCORRECT QUESTION :-

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Solve the value of x:

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

\blue\bigstarFORMULA 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\bigstarSOLUTION:-

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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}

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