Expand using suitable identities: [(4x/5) + (y/4)] [(4x/5) + (3y/4)]
Solution:
Given, [(4x/5) + (y/4)] [(4x/5) + (3y/4)]
Using standard identity: (x + a) (x + b) = x² + (a + b)x + ab
Here, a = (y/4) and b = (3y/4)
[(4x/5) + (y/4)] [(4x/5) + (3y/4)]
= [(4x/5)]² + {[(y/4) + (3y/4)](4x/5)} + [(y/4)×(3y/4)]
= (16x²/25) + [(y) ×(4x/5)] + (3y²/16)
= (16x²/25) + (4xy/5) + (3y²/16)
✦ Try This: Expand using suitable identities: [(2a/3) + (b/7)] [(2a/3) - (2b/9)]
Given, [(2a/3) + (b/7)] [(2a/3) - (2b/9)]
Using standard identity: (x + a) (x + b) = x² + (a + b)x + ab
[(2a/3) + (b/7)] [(2a/3) - (2b/9)] = [(2a/3)]² + {[(b/7) - (2b/9)](2a/3)} + [(2a/3)(- (2b/9))]
= 4a²/9 + [(-5b/63)(2a/3)] - 4ab/27
= 4a²/9 - 10ab/189 - 4ab/27
☛ Also Check: NCERT Solutions for Class 8 Maths Chapter 9
NCERT Exemplar Class 8 Maths Chapter 7 Problem 85(viii)
Expand using suitable identities: [(4x/5) + (y/4)] [(4x/5) + (3y/4)]
Summary:
Expanding [(4x/5) + (y/4)] [(4x/5) + (3y/4)] we get, (16x²/25) + (4xy/5) + (3y²/16)
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