\sqrt{ {3x}^{2} } + 4x - 7 \sqrt{3} = 0 to solve this quadriatic equation by factorization complete the follo

Question

 \sqrt{ {3x}^{2} }  + 4x - 7 \sqrt{3}  = 0
to solve this quadriatic equation
by factorization complete the following activity​

in progress 0
Kaylee 1 month 2021-08-14T06:40:23+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-08-14T06:41:38+00:00

    GIVEN :

    • A quadratic equation √3x² + 4x – 7√3 = 0.

    TO FIND :

    • Find x by using Factors method .

    SOLUTION :

     \bf  \implies \:  \sqrt{3}  {x}^{2}  + 4x - 7 \sqrt{3}  = 0

    • Now Splitting Middle term –

     \bf  \implies \:  \sqrt{3}  {x}^{2}  + 7x - 3x - 7 \sqrt{3}  = 0

     \bf  \implies \: x( \sqrt{3}x + 7)-  \sqrt{3}( \sqrt{3} x - 7 ) = 0

     \bf  \implies \: (x -  \sqrt{3}) ( \sqrt{3}x + 7) = 0

    (i)ㅤ

     \bf  \implies \: x -  \sqrt{3}= 0

     \bf  \implies \large{ \boxed{ \bf x  =   \sqrt{3}}}

    ㅤㅤ

    (ii)

     \bf  \implies \: \sqrt{3}x + 7 = 0

     \bf  \implies \: \sqrt{3}x  =  -  7

     \bf  \implies \: x  =  -  \dfrac{7}{ \sqrt{3} }

    ㅤㅤ

     \bf  \implies \: x  =  -  \dfrac{7}{ \sqrt{3} } \times  \dfrac{ \sqrt{3} }{ \sqrt{3} }

     \bf  \implies \large { \boxed{\bf x  =  -  \dfrac{7 \sqrt{3} }{3}}}

    Hence , x = √3 , x = (-7√3)/3 .

    0
    2021-08-14T06:41:42+00:00

    \purple{\mathbb{ANSWER}}

    \sf \sqrt3{x²} + 4x - 7\sqrt3 = 0

    {\implies}\sf \sqrt3x² - 3x + 7x - 7\sqrt3 = 0

    {\implies}\sf \sqrt3x(x - \sqrt3) + 7(x - 3\sqrt) = 0

    {\implies}\sf (\sqrt3x + 7)(x - \sqrt3) = 0

    {\therefore}~~\sf \red{x =  \frac{-7}{\sqrt3}}  ~~\: and  ~~\:\red {x = \sqrt3}

Leave an answer

Browse
Browse

18:9+8+9*3-7:3-1*13 = ? ( )