Find the nature of roots . If real roots exist find them :-(1).2×squa

1.	Find the nature of roots . If real roots exist find them :-(1).2×square -6×square +3=0
Answer» The given quadratic equation is\xa02x2\xa0- 6x + 3= 0Here, a = 2, b = -6, c = 3Therefore, discriminant = b2\xa0- 4ac{tex}= ( - 6 ) ^ { 2 } - 4 ( 2 ) ( 3 ) = 36 - 24{/tex}= 12 > 0So, the given quadratic equation has two distinct real rootsSolving the quadratic equation\xa0{tex}2 x ^ { 2 } - 6 x + 3 = 0{/tex} ,\xa0by the quadratic formula,\xa0{tex}x = \\frac { - b \\pm \\sqrt { b ^ { 2 } - 4 a c } } { 2 a }{/tex}we get {tex}= \\frac { - ( - 6 ) \\pm \\sqrt { 12 } } { 2 ( 2 ) } = \\frac { 6 \\pm 2 \\sqrt { 3 } } { 4 } = \\frac { 3 \\pm \\sqrt { 3 } } { 2 }{/tex}Therefore, the roots are\xa0{tex}\\frac { 3 \\pm \\sqrt { 3 } } { 2 } , \\text { i.e. } \\frac { 3 + \\sqrt { 3 } } { 2 } \\text { and } \\frac { 3 - \\sqrt { 3 } } { 2 }{/tex}\u200b\u200b\u200b\u200b\u200b\u200b

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