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If \(P^2-QR=10,\) \(Q^2+PR=10,\) \(R^2+PQ=10,\) and \(R\ne Q,\) what is the value of \(P^2+Q^2+R^2?\)

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Re: If \(P^2-QR=10,\) \(Q^2+PR=10,\) \(R^2+PQ=10,\) and \(R\ne Q,\) what is the value of \(P^2+Q^2+R^2?\)

by regor60 » Sun Mar 14, 2021 6:36 am
You're free to choose values for P,Q,and R that satisfy the equalities with the exception of R $$\ne$$ Q.

So, let P=Q

P^2 -PR = 10
P^2 + PR =10

from the first two equalities, therefore
P^2 = 10

this means that R must =0

Since P=Q, Q^2 also =10

So P^2+Q^2+R^2 =20,C
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