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

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

May 31, 2015

#1
+94558
+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 .........

May 31, 2015

#1
+94558
+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
#2
+95360
0

This one looks interesting CPhill :)

Jun 1, 2015