# Ex 9.3, 4 - Chapter 9 Class 8 Algebraic Expressions and Identities

Last updated at Dec. 24, 2018 by Teachoo

Last updated at Dec. 24, 2018 by Teachoo

Transcript

Ex 9.3, 4 (a) (a) Simplify 3x (4x β 5) + 3 and find its values for (i) x = 3 (ii) x = 1/2 3π₯ (4π₯β5)+3 = (3π₯Γ4π₯)β(3π₯Γ5)+3 = 12π₯^2β15π₯+3 (i) For π=π Putting π₯=3 in expression 12π₯^2β15π₯+3 = 12(3)^2β15(3)+3 = (12Γ9)β(15Γ3) +3 = 108β45 +3 = 108β42 = 66 (ii) For π=π/π Putting π₯=1/2 in expression 12π₯^2β15π₯+3 = 12(1/2)^2β15(1/2)+3 = 12(1/2Γ1/2)β 15/2 +3 = 12Γ(1/4)β15/2+3 = 3β15/2+3 = 6β15/2 = (6 Γ 2 β 15)/2 = (12 β 15)/2 = (βπ)/π Ex 9.3, 4 (b) (b) Simplify a (a2 + a + 1) + 5 and find its value for (i) a = 0, (ii) a = 1 (iii) a = β 1. π (π^2+π+1)+5 = (πΓπ^2 )+(πΓπ)+(πΓ1)+5 = π^3+π^2+π+5 (i) For π=π Putting π=0 in expression π^3+π^2+π+5 = (0)^3+(0)^2+0+5 = 0+0+0+5 = π (i) For π=π Putting π=1 in expression π^3+π^2+π+5 = γ(1)γ^3+ (1)^2+1+5 = 1+1+1+5 = π (iii) For π=βπ Putting π=β1 in expression π^3+π^2+π+5 = γ(β1)γ^3+ (β1)^2+(β1)+5 = (β1)^2Γ(β1)+(1)+(β1)+5 = 1Γ(β1)+1β1+5 = β1+1β1+5 = β2+6 = π

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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.