Answered

Westonci.ca is the ultimate Q&A platform, offering detailed and reliable answers from a knowledgeable community. Discover the answers you need from a community of experts ready to help you with their knowledge and experience in various fields. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

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]
We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.