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Let $$a$$ and $$b$$ be real numbers. One of the roots of $$x^3 + ax + b = 0$$ is $$1 + i \sqrt{3}.$$ Find $$a+b.$$

Apr 10, 2019

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Let  a and b be real numbers.
One of the roots of $$x^3 + ax + b = 0$$ is $$1 + i \sqrt{3}$$.
Find $$a+b$$.

$$\text{Let x_1=1+i \sqrt{3} } \\ \text{Let x_2=1-i \sqrt{3} }$$

$$\begin{array}{|rclrcl|} \hline &&\mathbf{ x^3+ax+b } \\ &=& (x-x_1)(x-x_2)(x-x_3) \\ &=&\left(x-(1+i \sqrt{3})\right) \left(x-(1-i \sqrt{3}) \right)(x-x_3) \\ &=&\left((x-1)-i \sqrt{3}\right) \left((x-1)+i \sqrt{3} \right)(x-x_3) \\ &=&\left((x-1)^2-(i \sqrt{3})^2\right) (x-x_3) \\ &=&\left(x^2-2x+1+3\right) (x-x_3) \\ &=& (x^2-2x+4 ) (x-x_3) \\ &=& x^3-x^2x_3-2x^2+2xx_3+4x-4x_3 \\ &=& x^3-\underbrace{(x_3+2)}_{=0}x^2+\underbrace{(2x_3+4)}_{=a}x\underbrace{-4x_3}_{=b} \\\\ && \begin{array}{|rclrcl|} \hline x_3+2 &=& 0 \\ \mathbf{x_3} &\mathbf{=} & \mathbf{-2} \\\\ a &=& 2x_3+4 \\ &=& 2\cdot(-2) + 4 \\ \mathbf{a} &\mathbf{=} & \mathbf{0} \\\\ b &=& -4x_3 \\ &=& -4\cdot(-2) \\ \mathbf{b} &\mathbf{=} & \mathbf{8} \\ \hline \end{array}\\ \hline \end{array}$$

$$a+b = 0+8 = 8$$

Apr 10, 2019