Let f(x) = x^4 + ax^3 + bx^2 + c be a polynom | vedclass

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(B) Given $f(x) = x^4 + ax^3 + bx^2 + c$ and $f(1) = -9$,we have $1 + a + b + c = -9$,so $a + b + c = -10$  $(1)$.The equation $4x^3 + 3ax^2 + 2bx = 0

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$20$

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(B) Given $f(x) = x^4 + ax^3 + bx^2 + c$ and $f(1) = -9$,we have $1 + a + b + c = -9$,so $a + b + c = -10$  $(1)$.The equation $4x^3 + 3ax^2 + 2bx = 0$ has $i\sqrt{3}$ as a root. Since the coefficients are real,$-i\sqrt{3}$ is also a root. The third root must be $0$ because the constant term is $0$.Thus,the roots are $0, i\sqrt{3}, -i\sqrt{3}$.The sum of roots taken two at a time is $\frac{2b}{4} = (i\sqrt{3})(-i\sqrt{3}) + 0 + 0 = 3$,so $b = 6$.The sum of roots is $-\frac{3a}{4} = 0 + i\sqrt{3} - i\sqrt{3} = 0$,so $a = 0$.Substituting $a = 0$ and $b = 6$ into $(1)$,we get $0 + 6 + c = -10$,so $c = -16$.Thus,$f(x) = x^4 + 6x^2 - 16 = 0$.Factoring,$(x^2 + 8)(x^2 - 2) = 0$,so the roots are $x^2 = -8$ and $x^2 = 2$.The roots are $\alpha_1 = i\sqrt{8}, \alpha_2 = -i\sqrt{8}, \alpha_3 = \sqrt{2}, \alpha_4 = -\sqrt{2}$.Calculating the sum of squares of magnitudes: $|\alpha_1|^2 + |\alpha_2|^2 + |\alpha_3|^2 + |\alpha_4|^2 = 8 + 8 + 2 + 2 = 20$.

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