Question

Find the root of the quadratic equations:

3x²-2√6x + 2 = 0


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Answer 1

Hello Brother

Topic:- Quardratic Equation

Equation:-

$\bold{ {3x}^{2} - 2 \sqrt{6x} + 2 = 0}$

$\bold{ \sqrt{3x} ( \sqrt{3x} - \sqrt{2} ) - \sqrt{2} ( \sqrt{3x} - \sqrt{2} ) = 0}$

$\bold{( \sqrt{3x} - \sqrt{2} )( \sqrt{3x} - \sqrt{2}) = 0}$

$\bold{ \sqrt{3x} = \sqrt{2} }$

$\bold{ \sqrt{3x} =- \sqrt{2}}$

$\bold{x = \frac{ \sqrt{2} }{ \sqrt{3} }}$

$\bold{other \: root}$

$\bold{ \sqrt{3x} - \sqrt{2} = 0}$

$\bold{ \sqrt{3x} =- \sqrt{2}}$


$\bold{x = \frac{ \sqrt{2} }{ \sqrt{3} }}$

Thanks!!!

Answer 2

given quadratic equation :

3 x^2 - 2√6x + 2 = 0

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step 1 Find discriminant :
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D = b^2 - 4ac

here , a = 3 , b = - 2√6 , c = 2 ( constant )

D = ( -2√6)^2 - 4 (3 ) (2)

D =24 - 24 = 0

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step2 Solve for 'x' by using quadratic formula :
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x = - b +- √D / 2 a

x = - (-2√6 ) +- √0 ) / 2 ( 3 )

x = 2√6 / 6 , 0 / 6

x = √6 / 3 , 0

x = √(2 × 3 ) /√3 × √3 , 0

x = √2 × √3 / √3 × √3 , 0

x =√(2 / 3 ) , 0

therefore , required roots are

x = √(2/3) , 0

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Your Answer : x = √(2/3) , 0
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