Video Transcript
If π, π, π, and π are
proportional, which of the following equals the square root of six π squared minus
nine π squared over six π squared minus nine π squared? Is it option (A) π over π, option
(B) π over π, option (C) π over π, or option (D) π over π?
Weβre told that π, π, π, and π
are proportional, so letβs begin by considering what this means. We can write this as π is to π as
π is to π. And we note that π and π are
known as the extremes and π and π are the means or the middle terms. Now we can also write this as π
over π is the same as π over π; that is, the proportion of π to π is equal to
that of π two π. And now if π is in the same
proportion to π as π is to π, we can say that for some constant π, π multiplied
by π is equal to π and π multiplied by π is equal to π.
And now squaring both sides of our
first equation, we have π squared π squared is equal to π squared. And squaring both sides of our
second equation gives us π squared π squared is equal to π squared. And so we have π squared in terms
of π squared and π squared in terms of π squared. So now letβs look at the given
expression. Inside our square root, we have six
π squared minus nine π squared over six π squared minus nine π squared. We could at this point simplify
this slightly by noting that in both the numerator and denominator, we have a common
factor of three. And taking this outside some
parentheses, we could eliminate this three. But in fact, we may as well leave
it as it is since, as weβll see, itβs going to cancel anyway.
The more important thing to note is
that we can substitute our expressions for π squared and π squared in terms of π
squared and π squared into the given expression. So now inside our square root, we
have six π squared minus nine π squared π squared over six π squared minus nine
π squared π squared. And now we see we have a common
factor of π squared in our numerator and π squared in our denominator. And taking these outside
parentheses, we see that we have a common factor of six minus nine π squared in the
numerator and the denominator.
So now dividing numerator and
denominator by six minus nine π squared, weβre left with the square root of π
squared over π squared. And recalling that the square root
of a fraction is the square root of the numerator over the square root of the
denominator, we have the square root of π squared over the square root of π
squared. And now recalling that the square
root of the square of the number is simply the number, weβre left with π over
π. If π, π, π, and π are
proportional then, the given expression is equal to option (D), and thatβs π over
π.