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H(x)=(x^2)+1 and K(x)=-(x^2)+4. If K(H)=0, what are the roots/solutions?

Guest May 31, 2015

Best Answer 

 #1
avatar+90180 
+10

If K(H) = 0 , this implies that.....

 

-((x^2+1)^2) + 4  = 0   multiply both sides by -1

 

(x^2+1)^2 - 4  = 0    simplify

 

(x^2 + 1)^2  = 4    and by the square root property, we have

 

x^2 + 1  = ± 2     subtract 1 from both sides

 

x^2  =  ± 2 - 1   ....so either.....

 

x^2  = -3    .......which leads to two  "non-real " solutions   ....  or .....

 

x^2  = 1    so that, using the square root property again,   x = ± 1

 

And these are the two "real" solutions.....check that they make the specified condition true .........

 

 

CPhill  May 31, 2015
 #1
avatar+90180 
+10
Best Answer

If K(H) = 0 , this implies that.....

 

-((x^2+1)^2) + 4  = 0   multiply both sides by -1

 

(x^2+1)^2 - 4  = 0    simplify

 

(x^2 + 1)^2  = 4    and by the square root property, we have

 

x^2 + 1  = ± 2     subtract 1 from both sides

 

x^2  =  ± 2 - 1   ....so either.....

 

x^2  = -3    .......which leads to two  "non-real " solutions   ....  or .....

 

x^2  = 1    so that, using the square root property again,   x = ± 1

 

And these are the two "real" solutions.....check that they make the specified condition true .........

 

 

CPhill  May 31, 2015
 #2
avatar+93691 
0

This one looks interesting CPhill :)

Melody  Jun 1, 2015

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