Find the squares of (y+3)
1 answer:
Answer:
Since we know that
(a+b)
2
=a
2
+2ab+b
2
(a−b)
2
=a
2
−2ab+b
2
Now,
(i)
(x+3y)
2
=x
2
+2×x×3y+(3y)
2
=x
2
+6xy+9y
2
(ii)
(2x−5y)
2
=(2x)
2
−2×2x×5y+(5y)
2
=4x
2
−20xy+25y
2
(iii)
(a+
5a
1
)
2
=a
2
+2×a×
2a
1
+(
2a
1
)
2
=a
2
+1+
4a
2
1
(iv)
(2a−1a)=(1a)
2
(1a)
2
=1a
2
Answer
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