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Show that the equation \( x^{9}-5 x^{5}+4 x^{4}+2 x^{2}+1=0 \) has atleast 6 imaginary solutions.Determine the number of positive and negative roots of the equation \( x^{3}-5 x^{3}-14 x^{3}=0 \) |
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Answer» P(x) = x9 – 5x5 + 4x4 + 2x2 + 1 (i) The number of sign changes in P(x) is 2. The number of positive roots is atmost 2. (ii) P(-x) = -x9 + 5x5 + 4x4 + 2x2 + 1. The number of sign changes in P(-x) is 1. The number of negative roots of P (x) is atmost 1. Since the difference of number of sign changes in P(-x) and number of negative zeros is even. P(x) has one negative root. (iii) 0 is not the zero of the polynomial P(x). So the number of real roots is almost 3. ∴ The number of imaginary roots at least 6. |
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