Question

Solve Root3x^2+11x+6root3 =0 (By quadratic formula)

Answer 1

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Given:}}}}$

$\star \: {\tt{3 {x}^{2} + 11x + 6 \sqrt{3} = 0}}$

${\sf{\underline{\blue{Now,}}}}</p><p>$

$\star \: {\sf{\orange{using \: quadratic \: formula}}}</p><p></p><p>$

${\sf{\blue{where \: \implies \: a = 3,}}}</p><p>$

${\sf{\blue{\implies \: b = 11,}}}</p><p>$

${\sf{\blue{\implies \: c = 6 \sqrt{3} ,}}}</p><p>$

${\tt{ \implies \: x = {b}^{2} - 4ac }}$

${\tt{ \implies \: x = {(11)}^{2} - 4(3)(6 \sqrt{3}) }}$

${\tt{ \implies \: x = {121- 12(6 \sqrt{3})}}}$

${\tt{ \implies \: x = {121- 72\sqrt{3}}}}$

${\tt{ \implies \: x = {121- 124.70 < 0}}}$

$\star \: {\sf{\underline{\blue{the \: eqaution \:has \: imaginary \: roots}}}}</p><p>$

${\bf{\red{\underline{conditions:}}}}$

${\sf{\green{if \: {b}^{2} - 4ac > 0 \: then \: two \: diffrerent \: real \: roots }}}$${\sf{\green{if \: {b}^{2} - 4ac = 0 \: then \: two \: same \: real \: roots }}}$ ${\sf{\green{if \: {b}^{2} - 4ac < 0 \: then \: two \: imaginary \: roots }}}$

Answer 2

Hey friend!

To solve √3x² + 11x + 6√3 = 0 by Quadratic formula

Answer:

x = $\frac{-11 \ - \ \sqrt{17} }{2}$ Or $\frac{-11 \ + \ \sqrt{17} }{2}$

Step-by-step explanation:

$According \ to \ the \ Quadratic \ Formula,\\\\We\ know\ that\\ \\if \ ax^2 + bx + c = 0\\\\then,\\\\x = \frac{- \ b \ + \ \sqrt{D}}{2a} \ or \ \frac{- \ b \ + \ \sqrt{D}}{2a}\\\\where \ D = b^2 - 4ac \\\\Given,\\\\ Equation \ is \ \sqrt{3} x^2 + 11x +6 \sqrt{3}\\ \star \ a = \sqrt{3}\\ \star \ b = 11\\\star \ c = 6\sqrt{3} \\\\ Either\\\implies \ x = \frac{- \ 11 \ + \ \sqrt{11^2\ -\ 6\ *\ 4\ * \sqrt{3}\ *\ \sqrt{3} }}{2\sqrt{3} }$

$\implies x = \frac{-11 \ + \ \sqrt{121\ -\ 24\ * \ 3 } }{2\sqrt{3} } \\\\\implies x = \frac{-11 \ + \ \sqrt{121 \ - \ 72} }{2\sqrt{3} } \\\\\implies x = \frac{-11 \ + \ \sqrt{51} }{2\sqrt{3} } \\\\ \\Or \\\implies x = \frac{- \ 11 \ - \ \sqrt{11^2\ -\ 6\ *\ 4\ *\ \sqrt{3}\ *\ \sqrt{3} }}{2\sqrt{3} }\\\\\implies x = \frac{-11 \ - \ \sqrt{121 \ - \ 72} }{2\sqrt{3} }\\ \\\implies x = \frac{-11 \ - \ \sqrt{51} }{2\sqrt{3} }\\ \\ \implies x = \frac{-11 \ - \ \sqrt{17} }{2}$

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