Mathematics, 05.11.2020 14:00, caliharris123
HELP HELP HELP 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
Answers: 2
![kx^2 +(5-k)x+3=0 ....(1)\\ \\ plug \: x = 3 \\ \\ k(3)^2 +(5-k) \times 3+3=0 \\ \\ 9k +15-3k+3=0 \\ \\ 6k +18=0 \\ \\ 6k = - 18 \\ \\ k = - \frac{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 = - \frac{1}{3} \\ \\ \purple{ other \: root = - \frac{1}{3} }](/tpl/images/0869/9186/b8c67.png)
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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...
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