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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(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: $|\alpha_1|^2 + |\alpha_2|^2 + |\alpha_3|^2 + |\alpha_4|^2 = 8 + 8 + 2 + 2 = 20$.

Similar Questions

If $\alpha+i \beta$ and $\gamma+i \delta$ are the roots of $x^2-(3-2 i) x-(2 i-2)=0$,where $i=\sqrt{-1}$,then $\alpha \gamma+\beta \delta$ is equal to :

If $|z - 25i| \le 15$,then $|\max \text{amp}(z) - \min \text{amp}(z)| = $

$(r, \theta)$ denotes $r(\cos \theta + i \sin \theta)$. If $x = (1, \alpha)$,$y = (1, \beta)$,$z = (1, \gamma)$ and $x + y + z = 0$,then $\sum \cos (2\alpha - \beta - \gamma) = $

$\cos (x + iy)$ is equal to

If $x = -5 + 2\sqrt{-4}$,then the value of the expression $x^4 + 9x^3 + 35x^2 - x + 4$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module