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

d. bc = 12 and ef = 16

step-by-step explanation:

we are given that δabc : δdef such that .

also, it is given that

so, from the options, we need that .

a.

b.

c.  

d.

as, we can see that only option d gives the same ratio as i.e. .

hence, the possible lengths are bc = 12 and ef = 16.