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Solve-for-a-b-and-c-a-b-c-1-2-i-abc-1-4-iii-ab-ac-bc-3-2-iv-




Question Number 80493 by TawaTawa last updated on 03/Feb/20
Solve for  a, b and c        a + b + c  =  (1/2)      ..... (i)         abc   =   − (1/4)     ...... (iii)        ab + ac + bc   =  (3/2)       ...... (iv)
$$\mathrm{Solve}\:\mathrm{for}\:\:\mathrm{a},\:\mathrm{b}\:\mathrm{and}\:\mathrm{c} \\ $$$$\:\:\:\:\:\:\mathrm{a}\:+\:\mathrm{b}\:+\:\mathrm{c}\:\:=\:\:\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:…..\:\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\:\:\:\mathrm{abc}\:\:\:=\:\:\:−\:\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\:\:……\:\left(\mathrm{iii}\right) \\ $$$$\:\:\:\:\:\:\mathrm{ab}\:+\:\mathrm{ac}\:+\:\mathrm{bc}\:\:\:=\:\:\frac{\mathrm{3}}{\mathrm{2}}\:\:\:\:\:\:\:……\:\left(\mathrm{iv}\right) \\ $$
Commented by mr W last updated on 03/Feb/20
a,b,c are the roots of  z^3 −(1/2)z^2 +(3/2)z+(1/4)=0
$${a},{b},{c}\:{are}\:{the}\:{roots}\:{of} \\ $$$${z}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{2}}{z}^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{2}}{z}+\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{0} \\ $$
Commented by TawaTawa last updated on 03/Feb/20
Sir help me how you derive the equation.  I will solve the polynimial
$$\mathrm{Sir}\:\mathrm{help}\:\mathrm{me}\:\mathrm{how}\:\mathrm{you}\:\mathrm{derive}\:\mathrm{the}\:\mathrm{equation}. \\ $$$$\mathrm{I}\:\mathrm{will}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{polynimial} \\ $$
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
$$\left({z}−{a}\right)\left({z}−{b}\right)\left({z}−{c}\right)= \\ $$$$={z}^{\mathrm{3}} −\left({a}+{b}+{c}\right){z}^{\mathrm{2}} +\left({ab}+{bc}+{ca}\right){z}−{abc}=\mathrm{0} \\ $$
Commented by TawaTawa last updated on 03/Feb/20
God bless you sir. I  appreciate.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\:\mathrm{appreciate}. \\ $$

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