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Find the zeroes of the following quadratic polynomials and verify the relationship betweenthe zeroes and the coefficients.(i) x² - 2x - 8(ii) 4s2- 4s+1(iii) 6x2 -3-7x(iv) 4u2 +8u(v) t2-15(vi) 3x2 -x-4 |
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Answer» Step-by-step EXPLANATION: (1) x² - 2x -8 x² - 4x + 2x - 8 x(x-4).+2(x-4) (x+2)(x-4)=0 x = -2 and 4 sum of zeroes(α+β) = -b/a -2 + 4 = -(-2)/1 2 = 2 product of zeroes (αβ) = c/a -2 × 4 = -8/1 -8 = -8 (2) 4s² - 4s + 1 4s² -2s -2s +1 2s(2s - 1).-1(2s - 1) (2s - 1)(2s - 1) s = 1/2 and 1/2 verification α+β = -b/a 1/2 + 1/2 = -(-4)/4 2/2 = 4/4 1 = 1 αβ = c/a 1/2 × 1/2 = 1/4 1/4 = 1/4 (3) 6x² -7X -3 6x² - 9x + 2x - 3 3X(2x-3).+1(2x-3) (3x + 1)(2x-3) = 0 x = -1/3 and 3/2 verification α + β = -b/a -1/3 + 3/2 = -(-7)/6 (-2+9)/6 = 7/6 7/6 = 7/6 αβ = c/a -1/3 × 3/2 = -3/6 -1/2 = -1/2 (4) 4u² + 8u 4u(u+2) = 0 u = 0 and -2 verification α + β = -b/a o + (-2) = -8/4 -2 = -2 αβ = c/a 0 × (-2) = 0/4 0 = 0 (5) t² - 15 = 0 t² = 15 t = ±√15 t = +√15 , -√15 verification α + β = -b/a -√15 + √15 = 0/1 0 = 0 αβ = c/a (-√15) × (√15) = -15/1 -15 = -15 (6) 3x² - x - 4 3x² -4x + 3x - 4 x(3x-4).+1(3x-4) (x+1)(3x-4) = 0 x = -1 , 4/3 verification α + β = -b/a -1 + 4/3 = -(-1)/3 (-3+4)/3 = 1/3 1/3 = 1/3 αβ = c/a -1 × 4/3 = -4/3 -4/3 = -4/3
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