Let f ( x ) = x 4 + a x 3 + b x 2 + c be a polynomial with real coefficients such that f ( 1 ) = - 9 . Suppose that i 3 is a root of the equation 4 x 3 + 3 a x 2 + 2 b x = 0 where i = - 1 . If 1 , 2 , 3 and 4 are all the roots of the equation

Question

Let $f (x) = x^{4} + a x^{3} + b x^{2} + c$ be a polynomial with real coefficients such that $f (1) = - 9$ . Suppose that $i \sqrt{3}$ is a root of the equation $4 x^{3} + 3 a x^{2} + 2 b x = 0$ where $i = \sqrt{- 1}$ . If $α_{1}, α_{2}, α_{3}$ and $α_{4}$ are all the roots of the equation $f (x) = 0$ , then $| α_{1} |^{2} + | α_{2} |^{2} + | α_{3} |^{2} + | α_{4} |^{2}$ is equal to _______. [2024]

Smart Achievers · Accepted Answer

(20)

$Given that f (1) = - 9 \Rightarrow 1 + a + b + c = - 9 . . . (i)$

$and 4 x^{3} + 3 a x^{2} + 2 b x = 0$

$\Rightarrow x = 0, or 4 x^{2} + 3 a x + 2 b = 0 . . . (i i)$

$\Rightarrow \sqrt{3} i and - \sqrt{3} i are roots of (ii)$

$\Rightarrow \sqrt{3} i - \sqrt{3} i = - \frac{3 a}{4}, \sqrt{3} i (- \sqrt{3} i) = \frac{2 b}{4}$

$\Rightarrow a = 0, b = 6, c = - 16 from (i)$

$\Rightarrow f (x) = 0 \Rightarrow x^{4} + 6 x^{2} - 16 = 0$

$\Rightarrow x^{2} = \frac{- 6 \pm \sqrt{36 + 64}}{2} = - 3 \pm 5 = 2, - 8$

$\Rightarrow x = - \sqrt{2}, \sqrt{2}, - 2 \sqrt{2} i, 2 \sqrt{2} i$

$\Rightarrow | α_{1} |^{2} + | α_{2} |^{2} + | α_{3} |^{2} + | α_{4} |^{2} = 20$

Solution

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Solution

Q. Let f(x)=x4+ax3+bx2+c be a polynomial with real coefficients such that f(1)=-9. Suppose that i3 is a root of the equation 4x3+3ax2+2bx=0 where i=-1. If α1,α2,α3 and α4 are all the roots of the equation f(x)=0, then |α1|2+|α2|2+|α3|2+|α4|2 is equal to _______. [2024]

Ans. (20) Given that f(1)=-9⇒1+a+b+c=-9 ...(i) and 4x3+3ax2+2bx=0 ⇒x=0, or 4x2+3ax+2b=0 ...(ii) ⇒3i and -3i are roots of (ii) ⇒3i-3i=-3a4, 3i(-3i)=2b4 ⇒a=0, b=6, c=-16 from (i) ⇒f(x)=0⇒x4+6x2-16=0 ⇒x2=-6±36+642=-3±5=2,-8 ⇒x=-2, 2, -22i, 22i ⇒|α1|2+|α2|2+|α3|2+|α4|2=20

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