We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
141
1
avatar

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
 #1
avatar+23181 
+3

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

 

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

 

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

 

\(a+b = 0+8 = 8\)

 

laugh

 Apr 10, 2019

4 Online Users