Answer :
Explanation:
The correct option is A
(
p
+
q
)
(
p
+
q
+
2
r
)
Given expression:
p
2
+
q
2
+
2
(
p
q
+
q
r
+
r
p
)
The above expression can be
rewritten as,
=
(
p
2
+
q
2
+
2
p
q
)
+
2
(
q
r
+
r
p
)
Now, applying the identity
(
a
+
b
)
2
=
a
2
+
2
a
b
+
b
2
,
we get
p
2
+
q
2
+
2
p
q
=
(
p
+
q
)
2
So,
(
p
2
+
q
2
+
2
p
q
)
+
2
(
q
r
+
r
p
)
=
(
p
+
q
)
2
+
2
r
(
p
+
q
)
=
(
p
+
q
)
(
p
+
q
)
+
2
r
(
p
+
q
)
Taking the common factor out,
we get,
=
(
p
+
q
)
(
p
+
q
+
2
r
)
Answer:
To prove that \( p^2 + q^2 + r^2 - pq - pr - qr \) is always non-negative, we can rewrite and analyze the expression.
Consider the expression:
\[ p^2 + q^2 + r^2 - pq - pr - qr \]
We can rearrange this expression by grouping terms involving \( p, q, \) and \( r \) in a more insightful way:
\[ \frac{1}{2} (2p^2 + 2q^2 + 2r^2 - 2pq - 2pr - 2qr) \]
This can be factored as:
\[ \frac{1}{2} \left[(p-q)^2 + (q-r)^2 + (r-p)^2\right] \]
To verify this factorization, let's expand the squared terms:
\[ (p-q)^2 = p^2 - 2pq + q^2 \]
\[ (q-r)^2 = q^2 - 2qr + r^2 \]
\[ (r-p)^2 = r^2 - 2rp + p^2 \]
Adding these together, we get:
\[ (p-q)^2 + (q-r)^2 + (r-p)^2 = (p^2 - 2pq + q^2) + (q^2 - 2qr + r^2) + (r^2 - 2rp + p^2) \]
\[ = p^2 + q^2 + r^2 - 2pq + q^2 - 2qr + r^2 - 2rp + p^2 \]
\[ = 2p^2 + 2q^2 + 2r^2 - 2pq - 2qr - 2rp \]
Now, dividing by 2, we obtain:
\[ p^2 + q^2 + r^2 - pq - pr - qr = \frac{1}{2} \left[(p-q)^2 + (q-r)^2 + (r-p)^2\right] \]
Since squares of real numbers are always non-negative, the sum of squares is also non-negative. Therefore,
\[ \frac{1}{2} \left[(p-q)^2 + (q-r)^2 + (r-p)^2\right] \geq 0 \]
Thus,
\[ p^2 + q^2 + r^2 - pq - pr - qr \geq 0 \]
Hence, the expression \( p^2 + q^2 + r^2 - pq - pr - qr \) is always non-negative.
Explanation:
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