Question

√(3x^-2)=2x-1 solve this as quadratic equation

Answer 1

Answer:

Required roots of the given equation are { 1 ± √( 1 + 8√3 ) } / 4 .

Step-by-step explanation:

Given equation to be solved : -

= > √{ 3x^( - 2 ) } = 2x - 1

= > { 3x^( - 2 ) }^( 1 / 2 ) = 2x - 1

= > √3 . x^( - 2 x 1 / 2 ) = 2x - 1

= > √3 . x^( - 1 ) = 2x - 1

= > √3 . 1 / x = 2x - 1

= > √3 / x = 2x - 1

= > √3 = x( 2x - 1 )

= > √3 = 2x^2 - x

= > 2x^2 - x - √3 = 0

By Using Quadratic Formula : -

= > x = [ - b ± √( b^2 - 4ac ) ] / 2a    [ where a , b and c are values which are put when compared with ax^2 + bx + c = 0 ]

= > x = [ - ( - 1 ) ± √{ ( - 1 )^2 - 4( 2 )( - √3 ) } ] / 2( 2 )

= > x = [ 1 ± √{ 1 + 8√3 } ] / 4

Hence the required roots of the given equation are { 1 ± √( 1 - 8√3 ) } / 4 .