#1**+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**+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