Question

What is the solution for this :
√3x2+4x-7√3=0​

Answer 1

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Question ::

What is the solution for this :

$\sqrt{3 {x}^{2} } + 4x - 7 \sqrt{3 } = 0$

$\: \bigstar \: \large \underline{ \boxed{ \bf \color{lime}{answer}}}$

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

$\: \\ \leadsto \underline{ \bf{by \: using \: determinant \: form}}$

$\: \: \: \: \: \leadsto \tt{ {b}^{2} - 4ac}$

$\: \: \: \: \leadsto \tt{ {(4)}^{2} - 4( \sqrt{3} )( - 7 \sqrt{3} )}$

$\: \: \: \: \leadsto \tt{16 + 28 \times 3}$

$\: \: \: \ \: \leadsto \tt{16 + 84}$

$\: \: \: \leadsto \underline{ \boxed{ \bf{100}}}$

$\: \: \: \star \underline { \boxed{ \bf{by \: using \: formula \: method}}}$

$\: \implies \displaystyle \tt{x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} }$

$\: \: \implies \displaystyle \tt{x = \frac{ - 4 \pm \sqrt{100} }{2 \times \sqrt{3} } }$

$\: \: \implies \displaystyle \tt{x = \frac{ - 4 \pm10}{2 \sqrt{3} }}$

$\: \: \implies \displaystyle \tt{x = \frac{ - 4 + 10}{2 \sqrt{3} } = \frac{6}{2 \sqrt{3} } }$

$\: \: \: \implies \displaystyle \tt{ \cancel \frac{6}{2} \frac{1}{ \sqrt{3} } = \frac{3}{ \sqrt{3} } }$

$\: \: \: \implies \displaystyle \tt{ \cancel \frac{3}{ \sqrt{3} } = \sqrt{3} }$

$\: \: \bigstar \underline { \boxed{ \bf{ \sqrt{3} \: is \: one \: root}}}$

$\: \: \implies \displaystyle \tt{x = \frac{ - 4 - 10}{2 \sqrt{3} } = \frac{ - 14}{2 \sqrt{3} } }$

$\: \: \implies \displaystyle \tt{ \cancel \frac{14}{2 \sqrt{3} } = \frac{7}{ \sqrt{3} } }$

$\: \: \bigstar \underline{ \boxed{ \bf{ \frac{7}{ \sqrt{3} } is \: the \: second \: root}}}$

$\: \: \red \bigstar \underline{ \boxed{ \sf{ \sqrt{3} \: and \: \frac{7}{ \sqrt{3} } is \: the \: roots \: of \: this \: quadratic \: equation}}}$