






Subscribe to our Youtube Channel - https://you.tube/teachoo
Last updated at Dec. 24, 2018 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, 5 Show that. (ii) (9πβ5π)^2+180ππ=(9π+5π)^2 Solving LHS (9πβ5π)^2+180ππ (πβπ)^2=π^2+π^2β2ππ Putting π = 9π & π = 5π (πβπ)^2=π^2+π^2β2ππ Putting π = 9π & π = 5π Solving RHS (9π+5π)^2 (π+π)^2=π^2+π^2+2ππ Putting π = 9π & π = 5π = (9π)^2+(5π)^2+2(9π)(5π) = 81π^2+25π^2+90ππ Thus LHS = RHS Hence proved Ex 9.5, 5 Show that. (iii) (4/3 πβ3/4 π)^2+2ππ=16/9 π^2+9/16 π^2 Solving LHS (4/3 πβ3/4 π)^2+2ππ (πβπ)^2=π^2+π^2β2ππ Putting π = 4/3 π & π = 3/4 π = (4/3 π)^2+(3/4 π)^2β2(4/3 π)(3/4 π)+2ππ = (4/3)^2Γπ^2+(3/4)^2Γπ^2β2ππ+2ππ = 16/9 π^2+9/16 π^2 = RHS Since LHS = RHS Hence proved Ex 9.5, 5 Show that. (iv) (4ππ+3π)^2β(4ππβ3π)^2=48ππ^2 Solving (πππ+ππ)^π (π+π)^2=π^2+π^2+2ππ Putting π = 4ππ & π = 3π (π+π)^2=π^2+π^2+2ππ Putting π = 4ππ & π = 3π = (4ππ)^2+(3π)^2+2(4ππ)(3π) = 16π^2 π^2+9π^2+24ππ^2 Solving (πππβππ)^π (πβπ)^2=π^2+π^2β2ππ Putting π = 4ππ & π = 3π = (4ππ)^2+(3π)^2β2(4ππ)(3π) = 16π^2 π^2+9π^2β24ππ^2 Solving LHS (4ππ+3π)^2β(4ππβ3π)^2 = (16π^2 π^2+9π^2+24ππ^2 )β(16π^2 π^2+9π^2β24ππ^2 ) = 16π^2 π^2+9π^2+24ππ^2β16π^2 π^2β9π^2+24ππ^2 = (16π^2 π^2β16π^2 π^2 )+(9π^2β9π^2 )+(24ππ^2+24ππ^2 ) = 0+0+48ππ^2 = 48ππ^2 = R.H.S Since LHS = RHS Hence proved Ex 9.5, 5 Show that. (v) (πβπ) (π+π)+(πβπ)(π+π)+(πβπ)(π+π)=0 Solving LHS (πβπ) (π+π)+(πβπ)(π+π)+(πβπ)(π+π) Using (π₯βπ¦)(π₯+π¦)=π₯^2+π¦^2 = (π^2βπ^2 )+(π^2βπ^2 )+(π^2βπ^2 ) = π^2βπ^2+π^2βπ^2+π^2βπ^2 = (π^2βπ^2 )+(βπ^2+π^2 )+(βπ^2+π^2 ) = 0+0+0 = 0 = R.H.S Since LHS = RHS Hence proved
About the Author