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inquire123 | Student | (Level 3) eNoter

Posted September 11, 2013 at 4:29 AM via web

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aruv | High School Teacher | (Level 2) Valedictorian

Posted September 11, 2013 at 4:41 AM (Answer #1)

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Given parabola

`y=x^2+2`          (i)

and line L posses through (0,1) , which is parallel to x-axis.

Thus equation of line L is

`y=1`                 (ii)

Parabola does not intersect line L  ,so  there are no points common in (i) and (ii).

In other words

`x^2+2=1`  has no real solution.

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aruv | High School Teacher | (Level 2) Valedictorian

Posted September 11, 2013 at 4:45 AM (Reply #1)

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Please add this  to answer to complete the answer

`x^2+2=k`

`x^2=k-2`

`x=+-sqrt(k-2)`

Thus  x has two distinct and real values if k -2 >0

i.e.

k > 2

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