Q10 of 38 Page 10

Let’s express the following expressions as a difference of two squares:

i.


ii.


iii. x


i. 391 × 409


= (400 – 9)(400 + 9)


[∵ a2 – b2 = (a – b)(a + b), here taking a = 400, b = 9]


= (400)2 – 92


ii. (4x + 3y)(2x – 3y)


= (3x + x + 3y)(3x – x – 3y)


= (3x + x + 3y)(3x – (x + 3y))


[∵ a2 – b2 = (a – b)(a + b), Here taking a = 3x, b = x + 3y]


= (3x)2 – (x + 3y)2


iii. x


= x.1




[∵ a2 – b2 = (a – b)(a + b), here taking ]


More from this chapter

All 38 →
8

andLet’s find out the square of the expressions given below using the above identities –

i.


ii.


iii.


iv.


9

andLet’s form a whole square of the expressions given below using the above identities-

i.


ii.


iii.


iv.


11

Let’s factorise :

i.


ii.


iii.


iv.


v.


vi.


vii.


viii.


ix.


x.


xi.


xii.


xiii.


12

Let’s find the product by formulae;

i.


ii.


iii.


iv.


v.