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Question Number 132475 by Study last updated on 14/Feb/21
prove that x_1 +x_2 =−(b/a) x_1 and x_2 are roots of ax^2 +bx+c=0
$${prove}\:{that}\:\:{x}_{\mathrm{1}} +{x}_{\mathrm{2}} =−\frac{{b}}{{a}} \\ $$$${x}_{\mathrm{1}} {and}\:{x}_{\mathrm{2}} {are}\:{roots}\:{of}\:{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$
Commented by MJS_new last updated on 14/Feb/21
it′s just the solution formula again. solve the quadratic and add the solutions. where′s the problem?
$$\mathrm{it}'\mathrm{s}\:\mathrm{just}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{formula}\:\mathrm{again}.\:\mathrm{solve} \\ $$$$\mathrm{the}\:\mathrm{quadratic}\:\mathrm{and}\:\mathrm{add}\:\mathrm{the}\:\mathrm{solutions}.\:\mathrm{where}'\mathrm{s} \\ $$$$\mathrm{the}\:\mathrm{problem}? \\ $$
Commented by EDWIN88 last updated on 14/Feb/21
may be the problem why x_1 +x_2 =−(b/a)
$$\mathrm{may}\:\mathrm{be}\:\mathrm{the}\:\mathrm{problem}\:\mathrm{why}\:\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} =−\frac{\mathrm{b}}{\mathrm{a}} \\ $$
Commented by MJS_new last updated on 14/Feb/21
as I said, solve by using the usual formula
$$\mathrm{as}\:\mathrm{I}\:\mathrm{said},\:\mathrm{solve}\:\mathrm{by}\:\mathrm{using}\:\mathrm{the}\:\mathrm{usual}\:\mathrm{formula} \\ $$
Commented by EDWIN88 last updated on 14/Feb/21
yes
$$\mathrm{yes} \\ $$
Answered by rs4089 last updated on 14/Feb/21
let x_1 and x_2 are roots of eq^n ax^2 +bx+c=0 we can writen as ax^2 +bx+c=a(x−x_1 )(x−x_1 ) compare cofficient of x both side... b=−a(x_1 +x_2 ) so x_1 +x_2 =((−b)/a)
$${let}\:{x}_{\mathrm{1}} \:{and}\:{x}_{\mathrm{2}} \:{are}\:{roots}\:{of}\:{eq}^{{n}} \:{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$${we}\:{can}\:{writen}\:{as}\:\:{ax}^{\mathrm{2}} +{bx}+{c}={a}\left({x}−{x}_{\mathrm{1}} \right)\left({x}−{x}_{\mathrm{1}} \right) \\ $$$${compare}\:{cofficient}\:{of}\:{x}\:{both}\:{side}… \\ $$$${b}=−{a}\left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} \right) \\ $$$${so}\:\:\:\:{x}_{\mathrm{1}} +{x}_{\mathrm{2}} =\frac{−{b}}{{a}} \\ $$$$ \\ $$
Answered by physicstutes last updated on 14/Feb/21
Or begin with the general solution to the quadratic: x = ((−b±(√(b^2 −4ac)))/(2a)) say, x_1 = ((−b+(√(b^2 −4ac)))/(2a)) and x_2 =((−b−(√(b^2 −4ac)))/(2a)) x_1 + x_2 = ((−b+(√(b^2 −4ac)))/(2a)) + ((−b−(√(b^2 −2ac)))/(2a)) = ((−b+(√(b^2 −4a4)) −b −(√(b^2 −4ac)))/(2a)) x_1 + x_2 = ((−b−b)/(2a)) = ((−b)/a) hence x_1 + x_2 =−(b/a) similarly it can be shown that x_1 x_2 = (c/a)
$$\mathrm{Or}\:\mathrm{begin}\:\mathrm{with}\:\mathrm{the}\:\mathrm{general}\:\mathrm{solution}\:\mathrm{to}\:\mathrm{the}\:\mathrm{quadratic}: \\ $$$$\:{x}\:=\:\frac{−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}} \\ $$$$\mathrm{say},\:{x}_{\mathrm{1}} \:=\:\frac{−{b}+\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}}\:\mathrm{and}\:{x}_{\mathrm{2}} \:=\frac{−{b}−\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}} \\ $$$${x}_{\mathrm{1}} +\:{x}_{\mathrm{2}} \:=\:\frac{−{b}+\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}}\:+\:\frac{−{b}−\sqrt{{b}^{\mathrm{2}} −\mathrm{2}{ac}}}{\mathrm{2}{a}}\:=\:\frac{−{b}+\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{a}\mathrm{4}}\:−{b}\:−\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}} \\ $$$$\:{x}_{\mathrm{1}} \:+\:{x}_{\mathrm{2}} \:=\:\frac{−{b}−{b}}{\mathrm{2}{a}}\:=\:\frac{−{b}}{{a}} \\ $$$$\mathrm{hence}\:{x}_{\mathrm{1}} \:+\:{x}_{\mathrm{2}} \:=−\frac{{b}}{{a}} \\ $$$$\mathrm{similarly}\:\mathrm{it}\:\mathrm{can}\:\mathrm{be}\:\mathrm{shown}\:\mathrm{that}\:\:{x}_{\mathrm{1}} {x}_{\mathrm{2}} \:=\:\frac{{c}}{{a}} \\ $$
Commented by MJS_new last updated on 14/Feb/21
that′s exactly what I recommended if we answer the easiest questions, people wil never learn anything. you can do all their thinking − at least all their homework − for them for all eternity.
$$\mathrm{that}'\mathrm{s}\:\mathrm{exactly}\:\mathrm{what}\:\mathrm{I}\:\mathrm{recommended} \\ $$$$\mathrm{if}\:\mathrm{we}\:\mathrm{answer}\:\mathrm{the}\:\mathrm{easiest}\:\mathrm{questions},\:\mathrm{people} \\ $$$$\mathrm{wil}\:\mathrm{never}\:\mathrm{learn}\:\mathrm{anything}.\:\mathrm{you}\:\mathrm{can}\:\mathrm{do}\:\mathrm{all} \\ $$$$\mathrm{their}\:\mathrm{thinking}\:−\:\mathrm{at}\:\mathrm{least}\:\mathrm{all}\:\mathrm{their}\:\mathrm{homework}\:− \\ $$$$\mathrm{for}\:\mathrm{them}\:\mathrm{for}\:\mathrm{all}\:\mathrm{eternity}. \\ $$

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