3xsqure+2root5-5=0 wheather the quadratic equation has …
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Sia ? 5 years, 6 months ago
The given equation is {tex} 3 x ^ { 2 } + 2 \sqrt { 5 } x - 5 = 0{/tex}
Here, a = 3, b = {tex}2 \sqrt { 5 }{/tex} and, c = -5
{tex}\therefore{/tex} D = b2 - 4ac = ({tex}2 \sqrt { 5 }{/tex})2 - 4 {tex}\times{/tex} 3 {tex}\times{/tex} - 5 = 20 + 60 = 80 > 0
So, the given equation has real roots, and are given by
{tex}\alpha = \frac { - b + \sqrt { D } } { 2 a } = \frac { - 2 \sqrt { 5 } + \sqrt { 80 } } { 2 \times 3 } = \frac { - 2 \sqrt { 5 } + 4 \sqrt { 5 } } { 6 } = \frac { 2 \sqrt { 5 } } { 6 } = \frac { \sqrt { 5 } } { 3 }{/tex}
and, {tex}\beta = \frac { - b - \sqrt { D } } { 2 a } = \frac { - 2 \sqrt { 5 } - \sqrt { 80 } } { 2 \times 3 } = \frac { - 2 \sqrt { 5 } - 4 \sqrt { 5 } } { 6 } = - \sqrt { 5 }{/tex}
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