Question

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Given kx^2 +(5-k)x+3=0, where k is not equal to zero, find k if one of the roots is 3, then find the other roots

Answer 1

k = - 3

Other root = - 1/3

Explanation:

`kx^2 +(5-k)x+3=0 ....(1)\\ \\ plug \: x = 3 \\ \\ k(3)^2 +(5-k) * 3+3=0 \\ \\ 9k +15-3k+3=0 \\ \\ 6k +18=0 \\ \\ 6k = - 18 \\ \\ k = - (18)/(6) \\ \\ \huge \red{ k = - 3 }\\ \\ plug \: k = - 3 \: in \: equation \: (1) \\ \\ - 3 {x}^(2) + [5 - ( - 3)] x + 3 = 0 \\ \\ - 3 {x}^(2) + [5 + 3] x + 3 = 0 \\ \\ - 3 {x}^(2) + 8x + 3 = 0 \\ \\ 3 {x}^(2) - 8x - 3 = 0 \\ \\ 3 {x}^(2) - 9x + x - 3 = 0 \\ \\ 3x(x - 3) + 1(x - 3) = 0 \\ \\ (x - 3)(3x + 1) = 0 \\ \\ x - 3 = 0 \: \: or \: \: 3x + 1 = 0 \\ \\ x = 3 \: \: or \: x = - (1)/(3) \\ \\ \purple{ other \: root = - (1)/(3) }`