Show that root (4t^2-t^4) less equal 2 if -2 less eq t les eq 2?

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You need to prove the inequality `sqrt(4t^2-t^4) =< 2` , hence, factoring out `t^2` under the square root, yields:

`sqrt(t^2(4 - t)^2) = sqrt(t^2)*sqrt(4 - t^2) =< 2`

You should come up with the following notation such that:

`sqrt(t^2) = a; sqrt(4 - t^2) = b`

Notice that `(a+b)^2 = a^2 + 2ab + b^2` , hence, substituting back `sqrt(t^2)` for a and `sqrt(4 - t^2)` for b yields:

`(sqrt(t^2) + sqrt(4 - t^2))^2 = t^2 + 2sqrt(t^2)*sqrt(4 - t^2) + 4 - t^2`

`(sqrt(t^2) + sqrt(4 - t^2))^2 = 2sqrt(t^2)*sqrt(4 - t^2) + 4`

Notice that `(sqrt(t^2) + sqrt(4 - t^2))^2 >= 0 => 2sqrt(t^2)*sqrt(4 - t^2) + 4 >= 0 => 2sqrt(t^2)*sqrt(4 - t^2) >= -4`

Dividing by 2 yields:

`sqrt(t^2)*sqrt(4 - t^2) >= -2 => sqrt(t^2)*sqrt(4 - t^2) =< 2`

**Hence, evaluating the inequality using the fact that the square of a number is positive yields that `sqrt(4t^2-t^4) =< 2` holds for `t in [-2,2].` **

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