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| 1. |
3xsqure+2root5-5=0 wheather the quadratic equation has real roots and if so find the roots |
| Answer» The given equation is {tex} 3 x ^ { 2 } + 2 \\sqrt { 5 } x - 5 = 0{/tex}Here, a = 3, b =\xa0{tex}2 \\sqrt { 5 }{/tex} and, c = -5{tex}\\therefore{/tex}\xa0D = b2 - 4ac = ({tex}2 \\sqrt { 5 }{/tex})2 - 4\xa0{tex}\\times{/tex} 3 {tex}\\times{/tex}\xa0- 5 = 20 + 60 = 80 > 0So, 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,\xa0{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} | |