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Question Number 80493 by TawaTawa last updated on 03/Feb/20

$$\mathrm{Solve}\mathrm{for}\mathrm{a},\mathrm{b}\mathrm{and}\mathrm{c}$$$$\mathrm{a}+\mathrm{b}+\mathrm{c}=\frac{1}{2}.....\left(\mathrm{i}\right)$$$$\mathrm{abc}=-\frac{1}{4}......\left(\mathrm{iii}\right)$$$$\mathrm{ab}+\mathrm{ac}+\mathrm{bc}=\frac{3}{2}......\left(\mathrm{iv}\right)$$

Commented by mr W last updated on 03/Feb/20

$$a,b,caretherootsof$$$$z^3-\frac{1}{2}{z}^2+\frac{3}{2}z+\frac{1}{4}=0$$

Commented by TawaTawa last updated on 03/Feb/20

$$\mathrm{Sir}\mathrm{help}\mathrm{me}\mathrm{how}\mathrm{you}\mathrm{derive}\mathrm{the}\mathrm{equation}.$$$$\mathrm{I}\mathrm{will}\mathrm{solve}\mathrm{the}\mathrm{polynimial}$$

Commented by mr W last updated on 03/Feb/20

$$(z-a)(z-b)(z-c)=$$$$=z^3-(a+b+c)z^2+(ab+bc+ca)z-abc=0$$

Commented by TawaTawa last updated on 03/Feb/20

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}.$$

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