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| 1. |
Find the value of k for which one root of quadratic equation kx2 -14x+8=0 is six times the other |
| Answer» Let one root = {tex}\\alpha{/tex}Other root = 6{tex}\\alpha{/tex}{tex}\\therefore{/tex}Sum of roots =\xa0{tex}\\alpha{/tex}\xa0+ 6{tex}\\alpha{/tex}\xa0=\xa0{tex}\\frac{14}{k}{/tex}or,\xa0{tex}7 \\alpha = \\frac { 14 } { k } \\text { or } \\alpha = \\frac { 2 } { k }{/tex} ...........(i)Product of roots\xa0{tex}\\alpha ( 6 \\alpha ) = \\frac { 8 } { k }{/tex}or,\xa0{tex}6 \\alpha ^ { 2 } = \\frac { 8 } { k }{/tex}..........(ii)Solving (i) and (ii),{tex}6 \\left( \\frac { 2 } { k } \\right) ^ { 2 } = \\frac { 8 } { k }{/tex}{tex}6 \\times \\frac { 4 } { k ^ { 2 } } = \\frac { 8 } { k }{/tex}or,\xa0{tex}\\frac { 3 } { k ^ { 2 } } = \\frac { 1 } { k }{/tex}or, 3k = k2or, 3k - k2\xa0= 0k[3-k] = 0{tex}\\therefore{/tex}\xa0k = 0 or k = 3k = 0 is not possibleHence, k = 3 | |