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If-the-difference-of-two-roots-of-x-2-lx-m-0-is-1-then-prove-that-l-2-4m-2-1-2m-2-

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Question Number 202459 by MATHEMATICSAM last updated on 27/Dec/23
If the difference of two roots of x^2 − lx + m = 0 is 1 then prove that l^2 + 4m^2 = (1 + 2m)^2 .
$$\mathrm{If}\:\mathrm{the}\:\mathrm{difference}\:\mathrm{of}\:\mathrm{two}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$${x}^{\mathrm{2}} \:−\:{lx}\:+\:{m}\:=\:\mathrm{0}\:\mathrm{is}\:\mathrm{1}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$${l}^{\mathrm{2}} \:+\:\mathrm{4}{m}^{\mathrm{2}} \:=\:\left(\mathrm{1}\:+\:\mathrm{2}{m}\right)^{\mathrm{2}} \:. \\ $$
Answered by aleks041103 last updated on 27/Dec/23
x_(1,2) =(1/2)(l±(√(l^2 −4m))) ⇒∣x_1 −x_2 ∣=(√(l^2 −4m))=1 ⇒l^2 −4m=1 ⇒l^2 =1+4m ⇒l^2 +4m^2 =1+4m+4m^2 =1+2(1)(2m)+(2m)^2 ⇒l^2 +4m^2 =(1+2m)^2
$${x}_{\mathrm{1},\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left({l}\pm\sqrt{{l}^{\mathrm{2}} −\mathrm{4}{m}}\right) \\ $$$$\Rightarrow\mid{x}_{\mathrm{1}} −{x}_{\mathrm{2}} \mid=\sqrt{{l}^{\mathrm{2}} −\mathrm{4}{m}}=\mathrm{1} \\ $$$$\Rightarrow{l}^{\mathrm{2}} −\mathrm{4}{m}=\mathrm{1} \\ $$$$\Rightarrow{l}^{\mathrm{2}} =\mathrm{1}+\mathrm{4}{m} \\ $$$$\Rightarrow{l}^{\mathrm{2}} +\mathrm{4}{m}^{\mathrm{2}} =\mathrm{1}+\mathrm{4}{m}+\mathrm{4}{m}^{\mathrm{2}} =\mathrm{1}+\mathrm{2}\left(\mathrm{1}\right)\left(\mathrm{2}{m}\right)+\left(\mathrm{2}{m}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{l}^{\mathrm{2}} +\mathrm{4}{m}^{\mathrm{2}} =\left(\mathrm{1}+\mathrm{2}{m}\right)^{\mathrm{2}} \\ $$
Answered by Rasheed.Sindhi last updated on 27/Dec/23
roots: α, α−1 (say) α+( α−1)=l ∧ α(α−1)=m l=2α−1 ∧ m=α(α−1) • l^2 + 4m^2 = (1 + 2m)^2 lhs: (2α−1)^2 +4(α(α−1))^2 =4α^4 −8α^3 +8α^2 −4α+1 rhs: (1+2α(α−1) )^2 =(1+2α^2 −2α)^2 =4α^4 −8α^3 +8α^2 −4α+1 ∵ lhs=rhs ∴ proved
$${roots}:\:\alpha,\:\alpha−\mathrm{1}\:\:\left({say}\right) \\ $$$$\alpha+\left(\:\alpha−\mathrm{1}\right)={l}\:\wedge\:\alpha\left(\alpha−\mathrm{1}\right)={m} \\ $$$${l}=\mathrm{2}\alpha−\mathrm{1}\:\:\wedge\:\:{m}=\alpha\left(\alpha−\mathrm{1}\right) \\ $$$$\bullet\:{l}^{\mathrm{2}} \:+\:\mathrm{4}{m}^{\mathrm{2}} \:=\:\left(\mathrm{1}\:+\:\mathrm{2}{m}\right)^{\mathrm{2}} \\ $$$${lhs}:\:\left(\mathrm{2}\alpha−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{4}\left(\alpha\left(\alpha−\mathrm{1}\right)\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:=\mathrm{4}\alpha^{\mathrm{4}} −\mathrm{8}\alpha^{\mathrm{3}} +\mathrm{8}\alpha^{\mathrm{2}} −\mathrm{4}\alpha+\mathrm{1} \\ $$$${rhs}:\:\left(\mathrm{1}+\mathrm{2}\alpha\left(\alpha−\mathrm{1}\right)\:\right)^{\mathrm{2}} =\left(\mathrm{1}+\mathrm{2}\alpha^{\mathrm{2}} −\mathrm{2}\alpha\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{4}\alpha^{\mathrm{4}} −\mathrm{8}\alpha^{\mathrm{3}} +\mathrm{8}\alpha^{\mathrm{2}} −\mathrm{4}\alpha+\mathrm{1} \\ $$$$\because\:{lhs}={rhs} \\ $$$$\therefore\:{proved} \\ $$
Answered by witcher3 last updated on 27/Dec/23
x_2 −x_1 =1 ⇒x_2 ^2 +x_1 ^2 −2x_1 x_2 =1 =x_1 ^2 +x_2 ^2 −2m=1 x_1 ^2 +x_2 ^2 =(x_1 +x_2 )^2 −2x_1 x_2 =l^2 −2m ⇒l^2 −2m−2m=1 ⇒l^2 +4m^2 =1+4m+4m^2 =(2m+1)^2
$$\mathrm{x}_{\mathrm{2}} −\mathrm{x}_{\mathrm{1}} =\mathrm{1} \\ $$$$\Rightarrow\mathrm{x}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{x}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{2x}_{\mathrm{1}} \mathrm{x}_{\mathrm{2}} =\mathrm{1} \\ $$$$=\mathrm{x}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{x}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{2m}=\mathrm{1} \\ $$$$\mathrm{x}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{x}_{\mathrm{2}} ^{\mathrm{2}} =\left(\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2x}_{\mathrm{1}} \mathrm{x}_{\mathrm{2}} =\mathrm{l}^{\mathrm{2}} −\mathrm{2m} \\ $$$$\Rightarrow\mathrm{l}^{\mathrm{2}} −\mathrm{2m}−\mathrm{2m}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{l}^{\mathrm{2}} +\mathrm{4m}^{\mathrm{2}} =\mathrm{1}+\mathrm{4m}+\mathrm{4m}^{\mathrm{2}} =\left(\mathrm{2m}+\mathrm{1}\right)^{\mathrm{2}} \\ $$

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