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Ex 9.5
Ex 9.5, 1 (ii) Important
Ex 9.5, 1 (iii)
Ex 9.5, 1 (iv)
Ex 9.5, 1 (v) Important
Ex 9.5, 1 (vi) Important
Ex 9.5, 1 (vii)
Ex 9.5, 1 (viii)
Ex 9.5, 1 (ix) Important
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) Important You are here
Ex 9.5, 2 (v)
Ex 9.5, 2 (vi) Important
Ex 9.5, 2 (vii) Important
Ex 9.5, 3 (i)
Ex 9.5, 3 (ii) Important
Ex 9.5, 3 (iii)
Ex 9.5, 3 (iv) Important
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Ex 9.5, 3 (vi)
Ex 9.5, 4 (i)
Ex 9.5, 4 (ii)
Ex 9.5, 4 (iii) Important
Ex 9.5, 4 (iv)
Ex 9.5, 4 (v) Important
Ex 9.5, 4 (vi)
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Ex 9.5, 5 (i)
Ex 9.5, 5 (ii)
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Ex 9.5, 5 (iv)
Ex 9.5, 5 (v) Important
Ex 9.5, 6 (i)
Ex 9.5, 6 (ii) Important
Ex 9.5, 6 (iii)
Ex 9.5, 6 (iv)
Ex 9.5, 6 (v) Important
Ex 9.5, 6 (vi)
Ex 9.5, 6 (vii) Important
Ex 9.5, 6 (viii)
Ex 9.5, 6 (ix) Important
Ex 9.5, 7 (i)
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Ex 9.5, 7 (iv) Important
Ex 9.5, 8 (i)
Ex 9.5, 8 (ii)
Ex 9.5, 8 (iii) Important
Ex 9.5, 8 (iv) Important
Last updated at March 22, 2023 by Teachoo
Ex 9.5, 2 Use the identity (π₯+π)(π₯+π)=π₯^2+(π+π)π₯+ππ to find the following products. (iv) (4π₯+5)(4π₯β1) (4π₯+5)(4π₯β1) = (4π₯+5) (4π₯+(β1)) (π¦+π)(π¦+π)=π¦^2+(π+π)π¦+ππ Putting π¦ = 4π₯ , π = 5 & π = β1 (π¦+π)(π¦+π)=π¦^2+(π+π)π¦+ππ Putting π¦ = 4π₯ , π = 5 & π = β1