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Question Number 159653 by MathsFan last updated on 19/Nov/21
The equation x^2 +2xp+q=0 and x^2 +2ax+b=0 have common roots, show that (q−b)^2 +4(a−p)(aq−pb)=0
$$\:{The}\:{equation}\:{x}^{\mathrm{2}} +\mathrm{2}{xp}+{q}=\mathrm{0} \\ $$$$\:{and}\:{x}^{\mathrm{2}} +\mathrm{2}{ax}+{b}=\mathrm{0}\:{have}\:{common} \\ $$$${roots},\:{show}\:{that}\:\left({q}−{b}\right)^{\mathrm{2}} +\mathrm{4}\left({a}−{p}\right)\left({aq}−{pb}\right)=\mathrm{0} \\ $$$$ \\ $$
Answered by Rasheed.Sindhi last updated on 19/Nov/21
The equations have common roots ⇒They′re equivalent: ((2p)/(2a))=(q/b) ⇒(p/a)=(q/b)⇒p=((aq)/b) (q−b)^2 +4(a−p)(aq−pb) =(q−b)^2 +4(a−((aq)/b))(aq−b(((aq)/b))) =(q−b)^2 +4(((ab−aq)/b))(((abq−abq)/b)) =(q−b)^2 ≠0 in general. Only equal to 0 when q=b
$$\mathcal{T}{he}\:{equations}\:{have}\:{common}\:{roots} \\ $$$$\Rightarrow\mathcal{T}{hey}'{re}\:{equivalent}: \\ $$$$\frac{\mathrm{2}{p}}{\mathrm{2}{a}}=\frac{{q}}{{b}} \\ $$$$\Rightarrow\frac{{p}}{{a}}=\frac{{q}}{{b}}\Rightarrow{p}=\frac{{aq}}{{b}} \\ $$$$\left({q}−{b}\right)^{\mathrm{2}} +\mathrm{4}\left({a}−{p}\right)\left({aq}−{pb}\right) \\ $$$$=\left({q}−{b}\right)^{\mathrm{2}} +\mathrm{4}\left({a}−\frac{{aq}}{{b}}\right)\left({aq}−{b}\left(\frac{{aq}}{{b}}\right)\right) \\ $$$$=\left({q}−{b}\right)^{\mathrm{2}} +\mathrm{4}\left(\frac{{ab}−{aq}}{{b}}\right)\left(\frac{\cancel{{abq}}−\cancel{{abq}}}{{b}}\right) \\ $$$$=\left({q}−{b}\right)^{\mathrm{2}} \neq\mathrm{0}\:{in}\:{general}. \\ $$$${Only}\:\:{equal}\:{to}\:\mathrm{0}\:{when}\:{q}={b} \\ $$
Commented by MathsFan last updated on 19/Nov/21
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Commented by mr W last updated on 19/Nov/21
please recheck sir: it must be q=b! otherwise there are no common roots! ⇒They′re equivalent: ((2p)/(2a))=(q/b)=((1 (coef. of x^2 term))/1) ⇒p=a and q=b example: x^2 +3x+4=0 and x^2 +6x+8=0 have no common roots, even (3/6)=(4/8).
$${please}\:{recheck}\:{sir}: \\ $$$${it}\:{must}\:{be}\:{q}={b}!\:{otherwise}\:{there}\:{are} \\ $$$${no}\:{common}\:{roots}! \\ $$$$\Rightarrow\mathcal{T}{hey}'{re}\:{equivalent}: \\ $$$$\frac{\mathrm{2}{p}}{\mathrm{2}{a}}=\frac{{q}}{{b}}=\frac{\mathrm{1}\:\left({coef}.\:{of}\:{x}^{\mathrm{2}} \:{term}\right)}{\mathrm{1}} \\ $$$$\Rightarrow{p}={a}\:{and}\:{q}={b} \\ $$$$ \\ $$$${example}: \\ $$$${x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{4}=\mathrm{0}\:{and}\:{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{8}=\mathrm{0}\:{have} \\ $$$${no}\:{common}\:{roots},\:{even}\:\frac{\mathrm{3}}{\mathrm{6}}=\frac{\mathrm{4}}{\mathrm{8}}. \\ $$
Commented by Rasheed.Sindhi last updated on 20/Nov/21
Right Sir,ThanX!
$$\mathcal{R}{ight}\:\mathcal{S}{ir},\mathcal{T}{han}\mathcal{X}! \\ $$

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