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| 1. |
If ratio of the roots of equation px2+qx+q=0 is a:b, prove that √a÷b + √b÷a + √q÷p=0. |
| Answer» Let the roots of the given equation be {tex}\\alpha {/tex}\xa0and {tex}\\beta{/tex}.Then, {tex}\\alpha {/tex}\xa0+ {tex}\\beta{/tex}\xa0= {tex}{{ - q} \\over p}{/tex} and {tex}\\alpha .\\beta = {q \\over p}{/tex}Given: {tex}{\\alpha \\over \\beta } = {a \\over b}{/tex}Now, {tex}\\sqrt {{a \\over b}} + \\sqrt {{b \\over a}} + \\sqrt {{q \\over p}} = 0{/tex}=> {tex}\\sqrt {{\\alpha \\over \\beta }} + \\sqrt {{\\beta \\over \\alpha }} + \\sqrt {\\alpha \\beta } = 0{/tex} [Since, {tex}{\\alpha \\over \\beta } = {a \\over b}{/tex}\xa0and {tex}\\alpha .\\beta = {q \\over p}{/tex}]=> {tex}{{\\sqrt \\alpha } \\over {\\sqrt \\beta }} + {{\\sqrt \\beta } \\over {\\sqrt \\alpha }} + \\sqrt {\\alpha \\beta } = 0{/tex}=> {tex}{{\\alpha + \\beta + \\alpha \\beta } \\over {\\sqrt {\\alpha \\beta } }} = 0{/tex}=> {tex}\\alpha + \\beta + \\alpha \\beta = 0{/tex}=> {tex}{{ - q} \\over p} + {q \\over p} = 0{/tex} [Since, {tex}\\alpha + \\beta = {{ - q} \\over p}{/tex}\xa0and {tex}\\alpha \\beta = {q \\over p}{/tex}]=> {tex}-q+q=0{/tex}=> 0 = 0Hence proved. | |