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| 1. |
Find the zeros of f(x)=x3-kx2+39x-28 if it is given that the zeroes are in a.p |
| Answer» Let\xa0{tex}\\begin{array}{l}\\mathrm\\alpha,\\mathrm\\beta\\;\\mathrm{and}\\;\\mathrm \\gamma\\;\\mathrm{be}\\;\\mathrm{the}\\;\\mathrm{zeroes}\\;\\mathrm{of}\\;\\mathrm f(\\mathrm x)\\\\\\end{array}{/tex}and\xa0the zeroes are in AP.Suppose\xa0{tex} \\alpha = a - d , \\beta = a \\text { and } \\gamma = a + d{/tex}\xa0f(x) =\xa0x3 - 12x2 + 39x - 28{tex} \\therefore \\quad \\alpha + \\beta + \\gamma = - \\left( \\frac { - 12 } { 1 } \\right){/tex}= 12 ---------(1)and,\xa0{tex} \\alpha \\beta \\gamma = \\left( \\frac { - 28 } { 1 } \\right){/tex}= - 28 ---------(2)From equation\xa0(1) , we havea - d + a + a + d = 123a = 12\xa0{tex}\\Rightarrow{/tex}\xa0a = 4.Now from equation (2), we have(a-d)a(a+d) = 28a(a2-d2) = 284(16-d2) = 284d2\xa0= 64 - 28 = 36d2 = 36d =\xa0{tex}\\pm 6{/tex}Case I:\xa0When a = 4 and d = +6 : In this case,{tex}\\alpha{/tex}\xa0= a - d = 4 - 6\xa0= - 2,{tex} \\beta{/tex}\xa0= a = 4 and {tex} \\gamma{/tex}\xa0= a + d = 10CASE II: When a = 4 and d = -6: In this case,{tex}\\alpha{/tex}\xa0= a - d = 4 - (- 6 ) = 10 ,{tex} \\beta{/tex} = a = 4 and {tex} \\gamma{/tex}\xa0= a + d = 4 - 6 = - 2Hence, in either case the zeroes\xa0of the given polynomial are 10,4 and - 2. | |