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| 1. |
X²+6x-(a²+2a-8)=0 |
| Answer» The given equation is\xa0{tex}x^2\xa0+\xa06x -\xa0(a^2 +\xa02a - 8) = 0{/tex}Comparing it with {tex}Ax^2\xa0+ Bx +C = 0,{/tex} we get{tex}A = 1,\\ B = 6\\\xa0and\\ C = -(a^2 +\xa02a - 8){/tex}{tex}\\therefore{/tex}\xa0{tex}D =B^2\xa0- 4AC ={/tex} (6)2\xa0- 4(1)(-(a2 +\xa02a - 8))= 36 +4a2\xa0+ 8a - 32= 4a2\xa0+ 8a\xa0+\xa04 = 4(a2\xa0+ 2a + 1) = 4(a + 1)2\xa0> 0So, the given equation has real roots, given by{tex}\\alpha = \\frac { - B + \\sqrt { D } } { 2 A } = \\frac { - 6 + \\sqrt { 4 ( a + 1 ) ^ { 2 } } } { 2 \\times 1 } = \\frac { - 6 + 2 ( a + 1 ) } { 2 }{/tex}\xa0{tex}= -3 + (a + 1) = a - 2{/tex}{tex}\\beta = \\frac { - B - \\sqrt { D } } { 2 A } = \\frac { - 6 - \\sqrt { 4 ( a + 1 ) ^ { 2 } } } { 2 \\times 1 } = \\frac { - 6 - 2 ( a + 1 ) } { 2 }{/tex}\xa0{tex}= -3 - (a + 1) = -a - 4 = -(a + 4){/tex}Hence,\xa0(a - 2)\xa0and -(a + 4)\xa0are the roots of the given equation. | |