Last updated at Sept. 8, 2021 by Teachoo

Transcript

Ex 9.5, 5 Show that. (i) (3π₯+7)^2β84π₯=(3π₯β7)^2 Solving LHS (3π₯+7)^2β84π₯ = (3π₯)^2+(7)^2+ 2(3π₯)(7)β84π₯ = 9π₯^2+49+42π₯β84π₯ = 9π₯^2+49β42π₯ (π+π)^2=π^2+π^2β2ππ Putting π = 3π₯ & π = 7 = (3π₯)^2+(7)^2+ 2(3π₯)(7)β84π₯ = 9π₯^2+49+42π₯β84π₯ = 9π₯^2+49β42π₯ Solving RHS (3π₯β7)^2 (πβπ)^2=π^2+π^2β2ππ Putting π = 3π₯ & π = 7 = (3π₯)^2+(7)^2β2(3π₯)(7) = (3^2Γπ₯^2 )+49β(2Γ3Γ7)π₯ = 9π₯^2+49β42π₯ Thus LHS = RHS Hence proved

Ex 9.5

Ex 9.5, 1 (i)
Important

Ex 9.5, 1 (ii)

Ex 9.5, 1 (iii)

Ex 9.5, 1 (iv)

Ex 9.5, 1 (v)

Ex 9.5, 1 (vi)

Ex 9.5, 1 (vii)

Ex 9.5, 1 (viii)

Ex 9.5, 1 (ix)

Ex 9.5, 1 (x)

Ex 9.5, 2 (i)

Ex 9.5, 2 (ii)

Ex 9.5, 2 (iii)

Ex 9.5, 2 (iv)

Ex 9.5, 2 (v)

Ex 9.5, 2 (vi)

Ex 9.5, 2 (vii)

Ex 9.5, 3 (i)

Ex 9.5, 3 (ii)

Ex 9.5, 3 (iii)

Ex 9.5, 3 (iv)

Ex 9.5, 3 (v)

Ex 9.5, 3 (vi)

Ex 9.5, 4 (i)

Ex 9.5, 4 (ii)

Ex 9.5, 4 (iii)

Ex 9.5, 4 (iv)

Ex 9.5, 4 (v)

Ex 9.5, 4 (vi)

Ex 9.5, 4 (vii)

Ex 9.5, 5 (i) Important You are here

Ex 9.5, 5 (ii)

Ex 9.5, 5 (iii)

Ex 9.5, 5 (iv)

Ex 9.5, 5 (v)

Ex 9.5, 6 (i) Important

Ex 9.5, 6 (ii)

Ex 9.5, 6 (iii)

Ex 9.5, 6 (iv)

Ex 9.5, 6 (v)

Ex 9.5, 6 (vi)

Ex 9.5, 6 (vii)

Ex 9.5, 6 (viii)

Ex 9.5, 6 (ix)

Ex 9.5, 7 (i)

Ex 9.5, 7 (ii)

Ex 9.5, 7 (iii)

Ex 9.5, 7 (iv)

Ex 9.5, 8 (i) Important

Ex 9.5, 8 (ii)

Ex 9.5, 8 (iii)

Ex 9.5, 8 (iv)

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.