Answered

Discover the best answers at Westonci.ca, where experts share their insights and knowledge with you. Get quick and reliable solutions to your questions from a community of experienced experts on our platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.

H(x)=(x^2)+1 and K(x)=-(x^2)+4. If K(H)=0, what are the roots/solutions?
a. ±i√3
b. ±i
c. ±2
d. ±1
e. ±1, ±i√3

Sagot :

kate200468
[tex]H(x)=(x^2)+1\ \ \ \ and\ \ \ \ K(x)=-(x^2)+4. \\\\K(H)=-(H^2)+4=-(x^2+1)^2+4=4-(x^2+1)^2=\\\\.\ \ \ =2^2-(x^2+1)^2=(2-x^2-1)(2+x^2+1)=(1-x^2)(3+x^2)=\\\\.\ \ \ =(1-x)(1+x)(x^2-3\cdot i^2)=(1-x)(1+x)(x- \sqrt{3} \cdot i)(x+ \sqrt{3} \cdot i)\\\\ K(H)=0\ \ \ \ \Leftrightarrow\ \ \ (1-x)(1+x)(x- \sqrt{3} \cdot i)(x+ \sqrt{3} \cdot i)=0\\\\x=1\ \ \ \ or\ \ \ \ x=-1\ \ \ \ or\ \ \ \ x=\sqrt{3} \cdot i\ \ \ \ or\ \ \ \ x=-\sqrt{3} \cdot i\\\\Ans.\ e.[/tex]
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.