Find the nature of roots . …
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Sia ? 6 years, 5 months ago
The given quadratic equation is
2x2 - 6x + 3= 0
Here, a = 2, b = -6, c = 3
Therefore, discriminant = b2 - 4ac
{tex}= ( - 6 ) ^ { 2 } - 4 ( 2 ) ( 3 ) = 36 - 24{/tex}
= 12 > 0
So, the given quadratic equation has two distinct real roots
Solving the quadratic equation {tex}2 x ^ { 2 } - 6 x + 3 = 0{/tex} , by the quadratic formula, {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 {tex}\frac { 3 \pm \sqrt { 3 } } { 2 } , \text { i.e. } \frac { 3 + \sqrt { 3 } } { 2 } \text { and } \frac { 3 - \sqrt { 3 } } { 2 }{/tex}
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