Kevin said that if the index of a radical is even and the radicand is positive, then the radical has two real roots. Do you agree with Kevin? Explain why or why not.

Yes Kevin is correct

Step-by-step explanation:

The index, x, of a radical ˣ√ is the numerical value of the root sought of the number located under (within) the radical sign

Therefore, when the index is even, we have, numbers for the radical given by 2x, therefore we have;

⁽²ˣ⁾√(a²ˣ)

Where a = The 2x root of the radicand

Therefore, we can write, a²ˣ = aˣ × aˣ

For which aˣ can be neqative but will still give  a positive value

Therefore, when the index is even, the roots can either be a positive or a negative real number, which are two real numbers, +a or -a

find the magnitude of the trig term  

csc

(

x

)

−

1

by taking the absolute value of the coefficient.

1

find the lower bound of the range.

tap for more

y

=

−

2

find the upper bound of the range.

tap for more

y

=

0

the range is  

y

≤

−

2

or  

y

≥

0

.

(

−

∞

,

−

2

]

∪

[

0

,

∞

)

{

y

|

y

≤

−

2

,

y

≥

0

}

step-by-step explanation: