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2/x^2-5/x+2=0Bye using the method of compeleting the square. |
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Answer» {tex}{2\\over x^2}{/tex}-{tex}{5\\over x}{/tex}+2=0{tex}{2-5x+2x^2\\over x^2}{/tex}=02-5x+2x2=02x2-5x+2=0 ------(1)\xa0Now first of all we will make coefficient of x2\xa0to 1So divide eq.(1) by 2 we getx2-{tex}{5\\over 2}{/tex}x+1=0Now add and subtract the square of half of the coefficient of x we getx2-{tex}{5\\over2}{/tex}x+{tex}({5\\over 4})^2 -({5\\over 4})^2{/tex}+1=0(x-{tex}{5\\over 4})^2{/tex}-{tex}{25\\over 16}{/tex}+1=0 [ by using identity (a-b)2=a2-2ab+b2](x-{tex}{5\\over 4})^2{/tex}-{tex}{25+16\\over 16}{/tex}=0(x-{tex}{5\\over 4})^2{/tex}-{tex}{9\\over 16}{/tex}=0(x-{tex}{5\\over 4})^2{/tex}={tex}{9\\over 16}{/tex}(x-{tex}{5\\over 4}{/tex})={tex}{\\pm \\sqrt{9 \\over 16}}{/tex}(x-{tex}{5\\over 4}){/tex}={tex}{\\pm 3\\over 4}{/tex}So x-{tex}{5\\over 4}{/tex}={tex}{3\\over 4}{/tex}\xa0or x-{tex}{5\\over 4}{/tex}={tex}{-3\\over 4}{/tex}x={tex}{3\\over 4} +{5\\over 4}{/tex} or x={tex}{-3\\over 4}+{5\\over 4}{/tex}x={tex}{3+5\\over 4}{/tex}\xa0or x={tex}{-3+5\\over 4}{/tex}x={tex}{8\\over 4}{/tex}\xa0or x={tex}{2\\over 4}{/tex}x=2 or x=\xa0{tex}{1\\over 2}{/tex} {tex}2x^2-5x+2=0{/tex}{tex}=> (\\sqrt 2x)^2-2\\times \\sqrt 2 \\times {5\\over 2\\sqrt2}x+2 + ({5\\over 2\\sqrt 2})^2-({5\\over 2\\sqrt 2})^2=0{/tex}{tex}=> (\\sqrt 2x)^2-2\\times \\sqrt 2 \\times {5\\over 2\\sqrt2}x + ({5\\over 2\\sqrt 2})^2+2 -{25\\over 8}=0{/tex}{tex}=> (\\sqrt 2x)^2-2\\times \\sqrt 2 \\times {5\\over 2\\sqrt2}x + ({5\\over 2\\sqrt 2})^2-{9\\over 8}=0{/tex}{tex}=> (\\sqrt 2x-{5\\over 2\\sqrt 2})^2-({3\\over 2\\sqrt 2})^2=0{/tex}{tex}=> (\\sqrt 2x-{5\\over 2\\sqrt 2} +{3\\over 2\\sqrt 2})(\\sqrt 2x-{5\\over 2\\sqrt 2} -{3\\over 2\\sqrt 2}){/tex}{tex}=> (\\sqrt 2x-{1\\over \\sqrt 2} )(\\sqrt 2x-{\\sqrt 2}){/tex} |
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