# Ex 2.4, 1 (Optional) - Class 10

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 2.4, 1 Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case: (i) 2x3 + x2 β 5x + 2; 1/2 , 1, β 2 p(x) = 2x3 + x2 β 5x + 2 Verifying zeroes At x = π/π p(1/2) = 2 (1/2)^3 + (1/2)^2 β 5 (1/2) + 2 = 1/4 + 1/4 β 5/2 + 2 = (1 + 1 β 10 + 8)/4 = 0/4 = 0 Since p(1/2) = 0 Hence, 1/2 is a zero of p(x) At x = π p(1) = 2(1)3 + (1)2 β 5(1) + 2 = 2 + 1 β 5 + 2 = 5 β 5 = 0 Since p(1) = 0 Hence, 1 is a zero of p(x) At x = β2 p(-2) = 2(-2)3 + (-2)2 β 5(-2) + 2 = 16 + 4 + 10 + 2 = β16 + 16 = 0 Since p(-2) = 0 Hence, β2 is a zero of P(x). Verifying relationship between zeroes and coefficients. For a cubic Polynomial p(x) = ax3 + bx2 + cx + d With zeroes Ξ±, π½ and Ξ³ We have Ξ± + π½ + Ξ³ = (βπ)/π Ξ±"π½" + π½Ξ³ + Ξ³Ξ± = π/π Ξ±"π½" Ξ³= (βπ)/π For p(x) = 2x3 + x2 β 5x + 2 a = 3, b = β5, c = β11 and d = β3 And zeroes are πΌ = 1/2, π½ = 1 and πΎ = β2 Now πΆ+ π· + πΈ = 1/2 + 1 β 2 = (1 + 2 β 4)/2 = (β1)/2 = (βπ)/π πΆπ·+ π·πΈ + πΈπΆ = (1/2) (1) + (1) (β2) + (β2)(1/2) = 1/2 β 2 β 1 = (1 β 4 β 2)/2 = (β5)/2 = π/π πΆπ·πΈ = 1/2 Γ 1 Γ β2 = (β2)/2 = (βπ)/π Hence, the relationship is verified Ex 2.4, 1 Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case: (ii) x3 - 4x2 + 5x β 2; 2, 1, 1 p(x) = x3 - 4x2 + 5x β 2 Verifying zeroes At x = 2 p(2) = (2)3 β 4 (2)2 + 5(2) β 2 = 8 β 4(4)+ 10 β 2 = 18 β 18 = 0 Since p(2) = 0 Hence, 2 is a zero of p(x) At x = π p(1) = (1)3 β 4(1)2 + 5(1) β 2 = 1 β 4 + 5 β 2 = 5 β 5 = 0 Since p(1) = 0 Hence, 1 is a zero of p(x) At x = π p(1) = (1)3 β 4(1)2 + 5(1) β 2 = 1 β 4 + 5 β 2 = 5 β 5 = 0 Since p(1) = 0 Hence, 1 is a zero of p(x) Verifying relationship between zeroes and coefficients. For p(x) = x3 β 4x2 + 5x β 2 a = 1, b = β4, c = 5 and d = β2 And Zeroes are πΌ = 2, π½ = 1 and πΎ = 1 πΆ+ π· + πΈ = 2 + 1 + 1 = 4 = (β(β4))/1 = (βπ)/π Hence, the relationship is verified

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.