101×99=(100+1)(100 1)=100^2 1^2 [Applying the identity: (a+b)(a b)=a^2 b^2] nbsp; 105^2=(100+5)^2=(100+5)(100+5) 14y^3+7xy=7y(2y^2+x) 3×5x^23=5x^2 nbs ...

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Standard VIII

Mathematics

Algebraic Identity: (x + y)^2

Match the fol...

Question

Match the followings

(100 + 5) (100 + 5)

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100^{2} - 1^{2}

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7 y (2 y^{2} + x)

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5 x^{2}

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Solution

$101 \times 99 = (100 + 1) (100 - 1) = 100^{2} - 1^{2}$
[Applying the identity: $(a + b) (a - b) = a^{2} - b^{2}$ ]

$105^{2} = (100 + 5)^{2} = (100 + 5) (100 + 5)$

$14 y^{3} + 7 x y = 7 y (2 y^{2} + x)$

$\frac{3 \times 5 x^{2}}{3} = 5 x^{2}$

Suggest Corrections

Match the followings

101×99=(100+1)(100−1)=1002−12 [Applying the identity: (a+b)(a−b)=a2−b2] 1052=(100+5)2=(100+5)(100+5) 14y3+7xy=7y(2y2+x) 3×5x23=5x2

$101 \times 99 = (100 + 1) (100 - 1) = 100^{2} - 1^{2}$
[Applying the identity: $(a + b) (a - b) = a^{2} - b^{2}$ ]

$105^{2} = (100 + 5)^{2} = (100 + 5) (100 + 5)$

$14 y^{3} + 7 x y = 7 y (2 y^{2} + x)$

$\frac{3 \times 5 x^{2}}{3} = 5 x^{2}$