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25x-30x-1=0 quadratic equation |
| Answer» Give,\xa025x2\xa0- 30x + 11 = 0 ...(i)On comparing Eq. (i) with ax2\xa0+ bx + c = 0, we geta = 25, b = - 30 and c = 11.{tex}\\because{/tex}\xa0{tex}\\alpha{/tex}\xa0=\xa0{tex}\\frac { - b + \\sqrt { b ^ { 2 } - 4 a c } } { 2 a }{/tex}and\xa0{tex}\\beta = \\frac { - b - \\sqrt { b ^ { 2 } - 4 a c } } { 2 a }{/tex}{tex}\\therefore{/tex}\xa0{tex}\\alpha = \\frac { 30 + \\sqrt { ( - 30 ) ^ { 2 } - 4 \\times 25 \\times 11 } } { 2 \\times 25 }{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\alpha = \\frac { 30 + \\sqrt { 900 - 1100 } } { 50 }{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\alpha = \\frac { 30 + \\sqrt { - 200 } } { 50 } \\Rightarrow \\alpha = \\frac { 30 + 10 i \\sqrt { 2 } } { 50 }{/tex}and\xa0{tex}\\beta = \\frac { 30 - \\sqrt { ( - 30 ) ^ { 2 } - 4 \\times 25 \\times 11 } } { 2 \\times 25 }{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\beta = \\frac { 30 - \\sqrt { 900 - 1100 } } { 50 }{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\beta = \\frac { 30 - \\sqrt { - 200 } } { 50 } \\Rightarrow \\beta = \\frac { 30 - 10 i \\sqrt { 2 } } { 50 }{/tex}{tex}\\therefore{/tex}\xa0{tex}\\alpha = \\frac { 3 } { 5 } + \\frac { \\sqrt { 2 } } { 5 } i , \\beta = \\frac { 3 } { 5 } - \\frac { \\sqrt { 2 } } { 5 } i{/tex}Also, b2\xa0- 4ac = (- 30)2\xa0- 4\xa0{tex}\\times{/tex}\xa025\xa0{tex}\\times{/tex}\xa011 = 900 - 1100= - 200 < 0Hence, the roots are complex conjugate. | |