Question

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

Answer 1

`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- √(3) \cdot i)(x+ √(3) \cdot i)\\\\ K(H)=0\ \ \ \ \Leftrightarrow\ \ \ (1-x)(1+x)(x- √(3) \cdot i)(x+ √(3) \cdot i)=0\\\\x=1\ \ \ \ or\ \ \ \ x=-1\ \ \ \ or\ \ \ \ x=√(3) \cdot i\ \ \ \ or\ \ \ \ x=-√(3) \cdot i\\\\Ans.\ e.`