We have, (a+b+c) ^ 2 = a ^ 2 + b ^ 2 + c ^ 2 +2ab+2bc+2caApplying the above formula, we get[Here, a=x, nbsp; b=2y, nbsp; c= 4z](x+2y+4z) ^ 2 =( x )^ 2 +( 2y ...

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Standard VII

Mathematics

(a-b)^2 Expansion and Visualisation

Solve: x+2y+...

Question

Solve:
$(x + 2 y + 4 z)^{2}$

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Solution

We have, ${(a + b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a$
Applying the above formula, we get
[Here, $a = x, b = 2 y, c = 4 z$ ]
${(x + 2 y + 4 z)}^{2} = (x)^{2} + (2 y)^{2} + (4 z)^{2} + (2 \times x \times 2 y) + (2 \times 2 y \times 4 z) + (2 \times 4 z \times x)$
$= x^{2} + 4 y^{2} + 16 z^{2} + 4 x y + 16 y z + 8 x z$

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Solve:(x+2y+4z)2

We have, (a+b+c)2=a2+b2+c2+2ab+2bc+2caApplying the above formula, we get[Here, a=x,b=2y,c=4z](x+2y+4z)2=(x)2+(2y)2+(4z)2+(2×x×2y)+(2×2y×4z)+(2×4z×x) =x2+4y2+16z2+4xy+16yz+8xz

Solve:
$(x + 2 y + 4 z)^{2}$