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Question Number 162165 by cortano last updated on 27/Dec/21
If the ratio of the roots of equation ax^2 +2bx+c=0 is same as the ratio of the roots of px^2 +2qx+r=0 where a,b,c,p ,r are non zero real numbers . Then the value of ((b^2 /q^2 ))((p/a))((r/c)) is equal to
$$\mathrm{If}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{equation} \\ $$$$\:{ax}^{\mathrm{2}} +\mathrm{2}{bx}+{c}=\mathrm{0}\:\mathrm{is}\:\mathrm{same}\:\mathrm{as}\:\mathrm{the}\:\mathrm{ratio} \\ $$$$\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:{px}^{\mathrm{2}} +\mathrm{2}{qx}+{r}=\mathrm{0}\:\mathrm{where} \\ $$$$\:{a},{b},{c},{p}\:,{r}\:\mathrm{are}\:\mathrm{non}\:\mathrm{zero}\:\mathrm{real}\:\mathrm{numbers} \\ $$$$.\:\mathrm{Then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\left(\frac{{b}^{\mathrm{2}} }{{q}^{\mathrm{2}} }\right)\left(\frac{{p}}{{a}}\right)\left(\frac{{r}}{{c}}\right)\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\: \\ $$
Commented by bobhans last updated on 27/Dec/21
{ ((ax^2 +2bx+c=0⇒(x_1 /x_2 )=λ)),((px^2 +2qx+r=0⇒(x_3 /x_4 )=λ)) :}⇒(x_1 /x_2 ) = (x_3 /x_4 ) ⇒(x_2 /x_4 )=(x_1 /x_3 ) { ((x_1 .x_2 = (c/a)⇒λx_2 ^2 = (c/a) ; x_1 +x_2 =−((2b)/a))),((x_3 .x_4 =(r/p)⇒λx_4 ^2 = (r/p) ; x_3 +x_4 =−((2q)/p))) :} { ((x_2 (1+λ)=−((2b)/a)⇒1+λ=−((2b)/(ax_2 )))),((x_4 (1+λ)=−((2q)/p) ⇒1+λ=−((2q)/(px_4 )))) :} ⇒((2b)/(ax_2 )) = ((2q)/(px_4 )) ; (x_2 /x_4 ) = ((bp)/(aq)) ⇒(x_1 /x_3 ) .(x_2 /x_4 ) = ((cp)/(ar)) ⇒(c/r)=((ax_1 x_2 )/(px_3 x_4 )) ∴ ((b^2 /q^2 ))((p/a))((r/c))= (((bp)/(aq)))((b/q))((1/(c/r))) = ((x_2 /x_4 ))(((ax_2 )/(px_4 )))(((px_3 x_4 )/(ax_1 x_2 )))= ((x_2 x_3 )/(x_1 x_4 )) = 1
$$\:\begin{cases}{\mathrm{ax}^{\mathrm{2}} +\mathrm{2bx}+\mathrm{c}=\mathrm{0}\Rightarrow\frac{\mathrm{x}_{\mathrm{1}} }{\mathrm{x}_{\mathrm{2}} }=\lambda}\\{\mathrm{px}^{\mathrm{2}} +\mathrm{2qx}+\mathrm{r}=\mathrm{0}\Rightarrow\frac{\mathrm{x}_{\mathrm{3}} }{\mathrm{x}_{\mathrm{4}} }=\lambda}\end{cases}\Rightarrow\frac{\mathrm{x}_{\mathrm{1}} }{\mathrm{x}_{\mathrm{2}} }\:=\:\frac{\mathrm{x}_{\mathrm{3}} }{\mathrm{x}_{\mathrm{4}} }\:\Rightarrow\frac{\mathrm{x}_{\mathrm{2}} }{\mathrm{x}_{\mathrm{4}} }=\frac{\mathrm{x}_{\mathrm{1}} }{\mathrm{x}_{\mathrm{3}} } \\ $$$$\:\begin{cases}{\mathrm{x}_{\mathrm{1}} .\mathrm{x}_{\mathrm{2}} =\:\frac{\mathrm{c}}{\mathrm{a}}\Rightarrow\lambda\mathrm{x}_{\mathrm{2}} ^{\mathrm{2}} \:=\:\frac{\mathrm{c}}{\mathrm{a}}\:;\:\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} =−\frac{\mathrm{2b}}{\mathrm{a}}}\\{\mathrm{x}_{\mathrm{3}} .\mathrm{x}_{\mathrm{4}} =\frac{\mathrm{r}}{\mathrm{p}}\Rightarrow\lambda\mathrm{x}_{\mathrm{4}} ^{\mathrm{2}} =\:\frac{\mathrm{r}}{\mathrm{p}}\:;\:\mathrm{x}_{\mathrm{3}} +\mathrm{x}_{\mathrm{4}} =−\frac{\mathrm{2q}}{\mathrm{p}}}\end{cases} \\ $$$$\:\begin{cases}{\mathrm{x}_{\mathrm{2}} \left(\mathrm{1}+\lambda\right)=−\frac{\mathrm{2b}}{\mathrm{a}}\Rightarrow\mathrm{1}+\lambda=−\frac{\mathrm{2b}}{\mathrm{ax}_{\mathrm{2}} }}\\{\mathrm{x}_{\mathrm{4}} \left(\mathrm{1}+\lambda\right)=−\frac{\mathrm{2q}}{\mathrm{p}}\:\Rightarrow\mathrm{1}+\lambda=−\frac{\mathrm{2q}}{\mathrm{px}_{\mathrm{4}} }}\end{cases} \\ $$$$\:\:\Rightarrow\frac{\mathrm{2b}}{\mathrm{ax}_{\mathrm{2}} }\:=\:\frac{\mathrm{2q}}{\mathrm{px}_{\mathrm{4}} }\:;\:\frac{\mathrm{x}_{\mathrm{2}} }{\mathrm{x}_{\mathrm{4}} }\:=\:\frac{\mathrm{bp}}{\mathrm{aq}} \\ $$$$\:\Rightarrow\frac{\mathrm{x}_{\mathrm{1}} }{\mathrm{x}_{\mathrm{3}} }\:.\frac{\mathrm{x}_{\mathrm{2}} }{\mathrm{x}_{\mathrm{4}} }\:=\:\frac{\mathrm{cp}}{\mathrm{ar}}\:\Rightarrow\frac{\mathrm{c}}{\mathrm{r}}=\frac{\mathrm{ax}_{\mathrm{1}} \mathrm{x}_{\mathrm{2}} }{\mathrm{px}_{\mathrm{3}} \mathrm{x}_{\mathrm{4}} } \\ $$$$\:\therefore\:\left(\frac{\mathrm{b}^{\mathrm{2}} }{\mathrm{q}^{\mathrm{2}} }\right)\left(\frac{\mathrm{p}}{\mathrm{a}}\right)\left(\frac{\mathrm{r}}{\mathrm{c}}\right)=\:\left(\frac{\mathrm{bp}}{\mathrm{aq}}\right)\left(\frac{\mathrm{b}}{\mathrm{q}}\right)\left(\frac{\mathrm{1}}{\mathrm{c}/\mathrm{r}}\right) \\ $$$$\:\:=\:\left(\frac{\mathrm{x}_{\mathrm{2}} }{\mathrm{x}_{\mathrm{4}} }\right)\left(\frac{\mathrm{ax}_{\mathrm{2}} }{\mathrm{px}_{\mathrm{4}} }\right)\left(\frac{\mathrm{px}_{\mathrm{3}} \mathrm{x}_{\mathrm{4}} }{\mathrm{ax}_{\mathrm{1}} \mathrm{x}_{\mathrm{2}} }\right)=\:\frac{\mathrm{x}_{\mathrm{2}} \mathrm{x}_{\mathrm{3}} }{\mathrm{x}_{\mathrm{1}} \mathrm{x}_{\mathrm{4}} }\:=\:\mathrm{1} \\ $$
Answered by Rasheed.Sindhi last updated on 27/Dec/21
Let the roots of ax^2 +2bx+c=0 are α mα. the roots of px^2 +2qx+r=0 are β , mβ •α+mα=−((2b)/a) , mα^2 =(c/a) α=−((2b)/((1+m)a)), m(−((2b)/((1+m)a)))^2 =(c/a) •β+mβ=−((2q)/p) , mβ^2 =(r/p) β=−((2q)/((1+m)p)),m(−((2q)/((1+m)p)))^2 =(r/p) (( m(−((2b)/((1+m)a)))^2 )/(m(−((2q)/((1+m)p)))^2 ))=((c/a)/(r/p)) (( ((4b^2 )/((1+m)^2 a^2 )))/((4q^2 )/((1+m)^2 p^2 )))=((cp)/(ar)) ((4b^2 )/a^2 )×(p^2 /(4q^2 ))=((cp)/(ar)) ((b^2 p)/(aq^2 ))=(c/r)⇒((b^2 /q^2 ))((p/a))((r/c))=(c/r)((r/c))=1 ((b^2 /q^2 ))((p/a))((r/c))=1
$${Let}\:{the}\:{roots}\:{of}\:\:{ax}^{\mathrm{2}} +\mathrm{2}{bx}+{c}=\mathrm{0} \\ $$$${are}\:\alpha\:\:{m}\alpha. \\ $$$${the}\:{roots}\:{of}\:{px}^{\mathrm{2}} +\mathrm{2}{qx}+{r}=\mathrm{0}\:{are} \\ $$$$\:\beta\:,\:{m}\beta \\ $$$$\bullet\alpha+{m}\alpha=−\frac{\mathrm{2}{b}}{{a}}\:\:,\:{m}\alpha^{\mathrm{2}} =\frac{{c}}{{a}} \\ $$$$\:\:\:\alpha=−\frac{\mathrm{2}{b}}{\left(\mathrm{1}+{m}\right){a}},\:{m}\left(−\frac{\mathrm{2}{b}}{\left(\mathrm{1}+{m}\right){a}}\right)^{\mathrm{2}} =\frac{{c}}{{a}} \\ $$$$\bullet\beta+{m}\beta=−\frac{\mathrm{2}{q}}{{p}}\:,\:{m}\beta^{\mathrm{2}} =\frac{{r}}{{p}} \\ $$$$\:\:\:\:\beta=−\frac{\mathrm{2}{q}}{\left(\mathrm{1}+{m}\right){p}},{m}\left(−\frac{\mathrm{2}{q}}{\left(\mathrm{1}+{m}\right){p}}\right)^{\mathrm{2}} =\frac{{r}}{{p}} \\ $$$$\frac{\:\cancel{{m}}\left(−\frac{\mathrm{2}{b}}{\left(\mathrm{1}+{m}\right){a}}\right)^{\mathrm{2}} }{\cancel{{m}}\left(−\frac{\mathrm{2}{q}}{\left(\mathrm{1}+{m}\right){p}}\right)^{\mathrm{2}} }=\frac{\frac{{c}}{{a}}}{\frac{{r}}{{p}}} \\ $$$$\frac{\:\frac{\mathrm{4}{b}^{\mathrm{2}} }{\left(\mathrm{1}+{m}\right)^{\mathrm{2}} {a}^{\mathrm{2}} }}{\frac{\mathrm{4}{q}^{\mathrm{2}} }{\left(\mathrm{1}+{m}\right)^{\mathrm{2}} {p}^{\mathrm{2}} }}=\frac{{cp}}{{ar}} \\ $$$$\frac{\mathrm{4}{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }×\frac{{p}^{\mathrm{2}} }{\mathrm{4}{q}^{\mathrm{2}} }=\frac{{cp}}{{ar}} \\ $$$$\frac{{b}^{\mathrm{2}} {p}}{{aq}^{\mathrm{2}} }=\frac{{c}}{{r}}\Rightarrow\left(\frac{{b}^{\mathrm{2}} }{{q}^{\mathrm{2}} }\right)\left(\frac{{p}}{{a}}\right)\left(\frac{{r}}{{c}}\right)=\frac{{c}}{{r}}\left(\frac{{r}}{{c}}\right)=\mathrm{1} \\ $$$$\left(\frac{{b}^{\mathrm{2}} }{{q}^{\mathrm{2}} }\right)\left(\frac{{p}}{{a}}\right)\left(\frac{{r}}{{c}}\right)=\mathrm{1} \\ $$
Answered by Ar Brandon last updated on 27/Dec/21
−((2b)/a)=−((2q)/p)⇒(b/a)=(q/p)⇒(b/q)=(a/p)=(c/r) (c/a)=(r/p)⇒(p/a)=(r/c) ((b^2 /q^2 ))((p/a))((r/c))=((c^2 /r^2 ))((r/c))((r/c))=1
$$−\frac{\mathrm{2}{b}}{{a}}=−\frac{\mathrm{2}{q}}{{p}}\Rightarrow\frac{{b}}{{a}}=\frac{{q}}{{p}}\Rightarrow\frac{{b}}{{q}}=\frac{{a}}{{p}}=\frac{{c}}{{r}} \\ $$$$\frac{{c}}{{a}}=\frac{{r}}{{p}}\Rightarrow\frac{{p}}{{a}}=\frac{{r}}{{c}} \\ $$$$\left(\frac{{b}^{\mathrm{2}} }{{q}^{\mathrm{2}} }\right)\left(\frac{{p}}{{a}}\right)\left(\frac{{r}}{{c}}\right)=\left(\frac{{c}^{\mathrm{2}} }{{r}^{\mathrm{2}} }\right)\left(\frac{{r}}{{c}}\right)\left(\frac{{r}}{{c}}\right)=\mathrm{1} \\ $$
Commented by Rasheed.Sindhi last updated on 27/Dec/21
−((2b)/a)=−((2q)/p) ? −((2b)/a) is sum o f the roots in eqn 1 −((2q)/p) is sum o f the roots in eqn 2 How these sums are equal?
$$−\frac{\mathrm{2}{b}}{{a}}=−\frac{\mathrm{2}{q}}{{p}}\:? \\ $$$$−\frac{\mathrm{2}{b}}{{a}}\:{is}\:{sum}\:{o}\:{f}\:{the}\:{roots}\:{in}\:{eqn}\:\mathrm{1} \\ $$$$−\frac{\mathrm{2}{q}}{{p}}\:{is}\:{sum}\:{o}\:{f}\:{the}\:{roots}\:{in}\:{eqn}\:\mathrm{2} \\ $$$${How}\:{these}\:{sums}\:{are}\:{equal}? \\ $$
Commented by Ar Brandon last updated on 28/Dec/21
Thank you, Sir �� But I think my friend Dwaipayan Shikari was the most outstanding amongst we, the young members. I'm missing him already ���� He's been offline for a very long time even on telegram. He really motivated me a lot.
Commented by Ar Brandon last updated on 28/Dec/21
I love the way you often design your text, Sir. E.g. ThanKs Sir! 🤭
$$\mathrm{I}\:\mathrm{love}\:\mathrm{the}\:\mathrm{way}\:\mathrm{you}\:\mathrm{often}\:\mathrm{design}\:\mathrm{your}\:\mathrm{text},\:\mathrm{Sir}. \\ $$$$\mathcal{E}.{g}.\:\mathbb{T}\mathrm{han}\mathbb{K}\mathrm{s}\:\mathcal{S}{ir}! \\ $$🤭
Commented by Rasheed.Sindhi last updated on 28/Dec/21
α & mα roots of equation1 β & mβ roots of equation 2 α+mα=−((2b)/a) β+mβ=−((2q)/p) −((2b)/a)=−((2q)/p)⇒α+mα=β+mβ ⇒α(1+m)=β(1+m) ⇒α=β (Not given)
$$\alpha\:\&\:{m}\alpha\:{roots}\:{of}\:{equation}\mathrm{1} \\ $$$$\beta\:\&\:{m}\beta\:{roots}\:{of}\:{equation}\:\mathrm{2} \\ $$$$\alpha+{m}\alpha=−\frac{\mathrm{2}{b}}{{a}} \\ $$$$\beta+{m}\beta=−\frac{\mathrm{2}{q}}{{p}} \\ $$$$−\frac{\mathrm{2}{b}}{{a}}=−\frac{\mathrm{2}{q}}{{p}}\Rightarrow\alpha+{m}\alpha=\beta+{m}\beta \\ $$$$\Rightarrow\alpha\left(\mathrm{1}+{m}\right)=\beta\left(\mathrm{1}+{m}\right) \\ $$$$\Rightarrow\alpha=\beta\:\left({Not}\:{given}\right) \\ $$
Commented by Ar Brandon last updated on 28/Dec/21
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Commented by Rasheed.Sindhi last updated on 28/Dec/21
𝛂r 𝛃randon sir,am I wrong?
$$\boldsymbol{\alpha}{r}\:\boldsymbol{\beta}{randon}\:{sir},{am}\:{I}\:{wrong}? \\ $$
Commented by Ar Brandon last updated on 28/Dec/21
I'm not good enough ����
Commented by Rasheed.Sindhi last updated on 28/Dec/21
You′re more than enough good at your early stage of your life! This I say on the basis of your sent posts. (I recall that you are one of some youngest forum-friends.I′m your opposite in age as well as knowledge.) May God brightens your future!
$${You}'{re}\:{more}\:{than}\:{enough}\:{good}\:{at} \\ $$$${your}\:{early}\:{stage}\:{of}\:{your}\:{life}!\:\mathcal{T}{his} \\ $$$$\:{I}\:{say}\:{on}\:{the}\:{basis}\:{of}\:{your}\:{sent}\:{posts}. \\ $$$$\left({I}\:{recall}\:{that}\:{you}\:{are}\:{one}\:{of}\:{some}\right. \\ $$$${youngest}\:{forum}-{friends}.{I}'{m}\:{your} \\ $$$$\left.{opposite}\:{in}\:{age}\:{as}\:{well}\:{as}\:{knowledge}.\right) \\ $$$${May}\:{God}\:{brightens}\:{your}\:{future}! \\ $$
Commented by Rasheed.Sindhi last updated on 28/Dec/21
No doubt Shikari was also outstanding and I also miss him.Thank you for conversation! BTW how did you bring this emoji in a comment written by forum keyboard?
$$\mathcal{N}{o}\:{doubt}\:{Shikari}\:{was}\:{also}\:{outstanding} \\ $$$${and}\:{I}\:{also}\:{miss}\:{him}.\mathcal{T}{hank}\:{you}\:{for} \\ $$$${conversation}! \\ $$$$\mathcal{BTW}\:{how}\:{did}\:{you}\:{bring}\:{this}\:{emoji} \\ $$$${in}\:{a}\:{comment}\:{written}\:{by}\:{forum} \\ $$$${keyboard}? \\ $$
Commented by Ar Brandon last updated on 28/Dec/21
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Commented by Ar Brandon last updated on 28/Dec/21
I clicked on line matrix then " insert plain text"
Commented by Rasheed.Sindhi last updated on 28/Dec/21
Thank you Ar Brandon! 👍👍👍😃
$$\mathcal{T}{hank}\:{you}\:{Ar}\:{Brandon}!\: \\ $$👍👍👍😃
Commented by Ar Brandon last updated on 28/Dec/21
🙃cool
🙃cool

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