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If sum of squares of zeros of polynomial x^2 + 8x+k is-40 then find the value of k |
| Answer» Let {tex}\\alpha,\\beta{/tex}\xa0be the zeros of the polynomial {tex}f(x)=x^2-8x+k{/tex}.Sum of zeroes =\xa0{tex} \\alpha + \\beta = - \\left( \\frac { - 8 } { 1 } \\right) = 8{/tex}\xa0and, Product of zeroes =\xa0{tex} \\alpha \\beta = \\frac { k } { 1 } = k{/tex}Now,\xa0{tex} \\alpha ^ { 2 } + \\beta ^ { 2 } = 40{/tex}{tex} \\Rightarrow \\alpha ^ { 2 } + \\beta ^ { 2 }+2 \\alpha\\beta-2 \\alpha\\beta= 40{/tex}{tex} \\Rightarrow \\quad ( \\alpha + \\beta ) ^ { 2 } - 2 \\alpha \\beta = 40{/tex}{tex} \\Rightarrow \\quad 8 ^ { 2 } - 2 k = 40{/tex}{tex} \\Rightarrow \\quad 2 k = 64 - 40 {/tex}{tex}\\Rightarrow 2 k = 24 {/tex}{tex}\\Rightarrow k = 12{/tex} | |