If and , where are two roots of the equation , , then is equal to: [2026]

Question

If $α = 1$ and $β = 1 + i \sqrt{2}$ , where $i = \sqrt{- 1}$ are two roots of the equation $x^{3} + a x^{2} + b x + c = 0$ , $a, b, c \in ℝ$ , then $\int_{- 1}^{1} (x^{3} + a x^{2} + b x + c) d x$ is equal to: [2026]

Smart Achievers · Accepted Answer

(3)

Roots of $x^{3} + a x^{2} + b c + c = 0$

FIGURE

Sum of roots $α + β + γ = - a$

$\Rightarrow a = - 3$

Product of roots $α β γ = - c$

$\Rightarrow 1 [1 - {(i \sqrt{2})}^{2}] = - c \Rightarrow c = - 3$

Sum of product of two roots

$α β + β γ + γ α = b$

$\Rightarrow 1 + i \sqrt{2} + 3 + 1 - i \sqrt{2} = b \Rightarrow b = 5$

Now, $\int_{- 1}^{1} (x^{3} + a x^{2} + b x + c) d x$

$= \int_{- 1}^{1} (x^{3} - 3 x^{2} + 5 x - 3) d x$

$= - 3 \times 2 \int_{- 1}^{1} (x^{2} + 1) d x$ $[∵ \int_{- 1}^{1} x^{3} d x = 0 . \int_{- 1}^{1} 5 x d x = 0 being odd functions]$

$= - 6 {[\frac{x^{3}}{3} + x]}_{0}^{1} = - 6 [\frac{1}{3} + 1]$

$= - 6 \times \frac{4}{3} = - 8$

Solution

Q.
If $α = 1$ and $β = 1 + i \sqrt{2}$ , where $i = \sqrt{- 1}$ are two roots of the equation $x^{3} + a x^{2} + b x + c = 0$ , $a, b, c \in ℝ$ , then $\int_{- 1}^{1} (x^{3} + a x^{2} + b x + c) d x$ is equal to: [2026]

1 –2

2 –4

3 –8

4 –10

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