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Question

Find the square of :

(i) x+3y

(ii) 2x5y

(iii) a+15a

(iv) 2a1a

(v) x2y+1

(vi) 3a2b5c

(vii) 2x+1x+1

(viii) 5x+2x

(ix) 2x3y+z

(x) x+1x1

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Solution

(i) (x+3y)2=(x)2+(3y)2+2×x×3y

=x2+9y2+6xy.

(ii) (2x5y)2=(2x)2+(5y)22×2x×5y

=4x2+25y220xy.

(iii) (a+15a)2=(a)2+(15a)2+2×a×15a

=a2+125a2+25.

(iv) (2a1a)2=(2a)2+(1a)22×2a×1a

=4a2+1a24.

(v) (x2y+1)2=(x)2+(2y)2+(1)2+2×x×2y+2×(2y)×1+2×1×x

=x2+4y2+14xy4y+2x.

(vi) (3a2b5c)2=(3a)2+(2b)2+(5c)2+2×3a×2b+2×(2b)(5c)+2×5c×3a

=9a2+4b2+25c212ab+20bc30ca.

(vii) (2x+1x+1)=(2x)2+(1x)2+(1)2+2×2x×1x+2×1x×1+2×1×2x

=4x2+1x2+1+4+2x+4x

=4x2+1x2+5+2x+4x.

(viii) (5x+2x)2=(5)2+(x)2+(2x)2+2×5+×(x)+2(x)×2x+2×2x×5

=25+x2+4x210x4+20x

=21+x2+4x210x+20x.

(ix) (2x3y+z)2=(2x)2+(3y)2+(z)2+2×2x×3y+2(3y)×z+2×z×2x

=4x2+9y2+z212xy6yz+4zx.

(x) (x+1x1)2=(x)2+(1x)2+(1)2+2×x×1x+2×1x×(1)+2(1)×x

=x2+1x2+1+22x2x

=x2+1x2+32x2x.


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