Question

# If one zero of the polynomial 3x2 - 8x - (2k + 1) is seven times the other, find the zeroes and the value of k.

Solution

## Let, f(x) = 3x^2 - 8x - (2k + 1) and α and β be its zeroes Here, a = 3  b = -8 c = -(2k + 1)  Given, ( 1 zero of polynomial is 7 times other,    α = 7β    -(i) Sum of roots, α + β = -b/a 7β + β = -(-8)/3    [Using (i)] 8β = 8/3 β = 1/3 Putting β = 1/3 in (i), we have α = 7 x 1/3 =  7/3 So, the zeroes are:-  α = 7/3 and β = 1/3 Now, Product of roots = αβ =  1/3 x 7/3 = 7/9 αβ = c/a c/a = 7/9 -(2k + 1)/ 3 = 7/9 2k + 1 = -7/3 2k = -7/3-1 = -10/3 k =-10/6 =  -5/3 So, value of k = -5/3

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