Question

Solve the above with clear steps.( quadratic equations).​

Answer 1

GIVEN :–

• √3 x² - √2 x + 3√3 = 0

TO FIND :–

• Value of 'x' = ?

SOLUTION :–

• Using formula –

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$\bf \implies{ \boxed{\bf \: x = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}}}$

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• Here –

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$\: \: \: \: \bf {\huge{.}} \: \: a = \sqrt{3}$

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$\: \: \: \: \bf {\huge{.}} \: \: b = - \sqrt{2}$

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$\: \: \: \: \bf {\huge{.}} \: \: c = 3\sqrt{3}$

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• Put the values –

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$\bf \implies x = \dfrac{ - ( - \sqrt{2} ) \pm \sqrt{ {( - \sqrt{2}) }^{2} - 4( \sqrt{3})(3 \sqrt{3}) } }{2 \sqrt{3} }$

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$\bf \implies x = \dfrac{ \sqrt{2} \pm \sqrt{ 2 - 4( \sqrt{3})(3 \sqrt{3}) } }{2 \sqrt{3} }$

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$\bf \implies x = \dfrac{ \sqrt{2} \pm \sqrt{ 2 - 12(3)} }{2 \sqrt{3} }$

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$\bf \implies x = \dfrac{ \sqrt{2} \pm \sqrt{ 2 - 36} }{2 \sqrt{3} }$

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$\bf \implies x = \dfrac{ \sqrt{2} \pm \sqrt{ - 34} }{2 \sqrt{3} }$

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$\bf \implies x = \dfrac{1 \pm \sqrt{ - 17} }{ \sqrt{6} }$

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$\bf \implies x = \dfrac{1 \pm \sqrt{17} \sqrt{ - 1} }{ \sqrt{6} }$

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$\bf \implies x = \dfrac{1 \pm \sqrt{17} \iota }{ \sqrt{6}} \: \: \: \:\:\: \: \{ \: \: \because \: \: \sqrt{ - 1} = \iota \}$

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• Positive (+) sign :–

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$\bf \implies \large{ \boxed{ \bf x = \dfrac{1 + \sqrt{17} \iota }{ \sqrt{6}}}}$

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• Negetive (-) sign :–

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$\bf \implies \large{ \boxed{ \bf x = \dfrac{1 - \sqrt{17} \iota }{ \sqrt{6}}}}$