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| 1. |
Find the roots of following equation1÷x+4 _ 1÷x_7= 11÷30 |
| Answer» {tex}\\frac { 1 } { x + 4 } - \\frac { 1 } { x - 7 } = \\frac { 11 } { 30 }{/tex} where {tex}x \\neq - 4,7{/tex}{tex}\\Rightarrow \\frac { ( x - 7 ) - ( x + 4 ) } { ( x + 4 ) ( x - 7 ) } = \\frac { 11 } { 30 }{/tex}{tex}\\Rightarrow \\frac { - 11 } { ( x + 4 ) ( x - 7 ) } = \\frac { 11 } { 30 }{/tex}{tex}\\Rightarrow{/tex}x2 - 7x + 4x - 28 = -30{tex}\\Rightarrow{/tex}x2 - 3x + 2= 0Comparing equation x2 - 3x + 2 = 0 with general form ax2 + bx + c = 0,We get\xa0a = 1, b = -3 and c = 2Using quadratic formula {tex}x = {-b \\pm \\sqrt{b^2-4ac} \\over 2a}{/tex}to solve equation,{tex}x = \\frac { 3 \\pm \\sqrt { ( - 3 ) ^ { 2 } - 4 ( 1 ) ( 2 ) } } { 2 \\times 1 }{/tex}{tex}\\Rightarrow x = \\frac { 3 \\pm \\sqrt { 1 } } { 2 }{/tex}\xa0{tex}\\Rightarrow x = \\frac { 3 + \\sqrt { 1 } } { 2 } , \\frac { 3 - \\sqrt { 1 } } { 2 }{/tex}{tex}\\Rightarrow{/tex}x = 2, 1 | |