2/x^2-5/x+2=0 Bye using the method of …

{tex}{2\over x^2}{/tex}-{tex}{5\over x}{/tex}+2=0

{tex}{2-5x+2x^2\over x^2}{/tex}=0

2-5x+2x²=0

2x²-5x+2=0 ------(1) 

Now first of all we will make coefficient of x² to 1

So divide eq.(1) by 2 we get

x²-{tex}{5\over 2}{/tex}x+1=0

Now add and subtract the square of half of the coefficient of x we get

x²-{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} or 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} or x={tex}{-3+5\over 4}{/tex}

x={tex}{8\over 4}{/tex} or  x={tex}{2\over 4}{/tex}

x=2 or x= {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}