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If one zero of the polynomial, (a⁴+4)x² + 13x + 4a, is reciprocal of the othe,Find \' a\'

Answer» Let one root =\xa0{tex}\\alpha {/tex}Then second root =\xa0{tex}1\\over \\alpha{/tex}We know\xa0Product of roots =\xa0{tex}c\\over a{/tex}{tex}=> \\alpha\\times {1\\over \\alpha}= {4a\\over a^2+4}{/tex}=>\xa0{tex}a^2+4=4a{/tex}{tex}=> a^2-4a+4= 0{/tex}{tex}=> (a-2)^2= 0{/tex}=> a-2= 0=> a = 2


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