miymiy1737
Answered

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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

Sagot :

[tex]H(x)=x^2+1;\ K(x)=-x^2+4\\\\K(H)=K(x^2+1)=-(x^2+1)^2+4=0\\\\(x^2+1)^2=4\iff x^2+1=\pm\sqrt4\\\\x^2+1=-2\ or\ x^2+1=2\ \ \ \ |subtract\ 1\ from\ both\ sides\\\\x^2=-3\ or\ x^2=1\\\\\boxed{x=\pm i\sqrt3\ or\ x=\pm1}[/tex]
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