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| 1. |
X2 _\xa02(a2+ b2)x + (a2_\xa0b2)2\xa0= 0Find the roots?\xa0 |
| Answer» {tex}{x^2} - 2\\left( {{a^2} + {b^2}} \\right) + {\\left( {{a^2} - {b^2}} \\right)^2} = 0{/tex}Using {tex}x = {-b \\pm \\sqrt{b^2-4ac} \\over 2a}{/tex}{tex}x = {{2\\left( {{a^2} + {b^2}} \\right) \\pm \\sqrt {{2^2}.{{\\left( {{a^2} + {b^2}} \\right)}^2} - 4 \\times 1 \\times {{\\left( {{a^2} - {b^2}} \\right)}^2}} } \\over {2 \\times 1}}{/tex}{tex}x = {{2\\left( {{a^2} + {b^2}} \\right) \\pm \\sqrt {{{\\left\\{ {4\\left( {{a^2} + {b^2}} \\right)} \\right\\}}^2} - 4{{\\left( {{a^2} - {b^2}} \\right)}^2}} } \\over 2}{/tex}{tex}x = {{2\\left[ {\\left( {{a^2} + {b^2}} \\right) \\pm 2\\sqrt {{{\\left( {{a^2} + {b^2}} \\right)}^2} - {{\\left( {{a^2} - {b^2}} \\right)}^2}} } \\right]} \\over 2}{/tex}{tex}x = {a^2} + {b^2} \\pm \\sqrt {\\left( {{a^2} + {b^2} + {a^2} - {b^2}} \\right)\\left( {{a^2} + {b^2} - {a^2} + {b^2}} \\right)} {/tex}{tex}x = {a^2} + {b^2} \\pm \\sqrt {\\left( {2{a^2}} \\right)\\left( {2{b^2}} \\right)} {/tex}{tex}x = {a^2} + {b^2} \\pm 2{a}{b}{/tex}Taking positive sign Taking negative sign{tex}x = {a^2} + {b^2} + 2{a}{b}{/tex}{tex}x = {a^2} + {b^2} - 2{a}{b}{/tex}{tex}x = {\\left( {a + b} \\right)^2}{/tex}{tex}x = {\\left( {a - b} \\right)^2}{/tex} | |