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find the zeroes of the polynomial f(x) = `x^3`-`12x^2` +39x -28 , if the zeroes are in A.P |
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Answer» Let the roots of the given equation `x^3-12x^2+39x - 28 = 0` are, `a-d, a and a+d` as they are in A.P. Then, sum of roots, `a-d+a+a+d = -(-12) ` `=> 3a = 12 => a= 4` Product in pairs of the roots. `a(a-d)+a(a+d)+(a-d)(a+d) = 39` `=>a(a-d+a+d))+a^2-d^2 = 39` `=>3a^2 - d^2 = 39` `=>3(4)^2 -39 = d^2` `=>d = +-3` If `d = 3`, roots are `1,4 and 7`. If `d = -3`, roots are `7,4 and 1`. |
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