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If-and-are-the-roots-of-ax-2-bx-c-0-and-k-and-k-are-the-roots-of-lx-2-mx-n-0-then-prove-that-k-1-2-b-a-m-l-

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Question Number 202114 by MATHEMATICSAM last updated on 21/Dec/23
If α and β are the roots of ax^2 + bx + c = 0 and α + k and β + k are the roots of lx^2 + mx + n = 0 then prove that k = (1/2)((b/a) − (m/l)).
$$\mathrm{If}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$${ax}^{\mathrm{2}} \:+\:{bx}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{and}\:\alpha\:+\:{k}\:\mathrm{and}\:\beta\:+\:{k}\:\:\mathrm{are}\: \\ $$$$\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:{lx}^{\mathrm{2}} \:+\:{mx}\:+\:{n}\:=\:\mathrm{0}\:\mathrm{then}\:\mathrm{prove} \\ $$$$\mathrm{that}\:{k}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{b}}{{a}}\:−\:\frac{{m}}{{l}}\right). \\ $$
Answered by aleks041103 last updated on 21/Dec/23
ax^2 +bx+c=0⇔x=α,β ⇒x^2 +(b/a)x+(c/a)=0⇔x=α,β lx^2 +mx+n=0 ⇔ x=α+k,β+k ⇒l(x+k)^2 +m(x+k)+n=0⇔x=α,β ⇒l(x^2 +2xk+k^2 )+mx+mk+n=0 lx^2 +(2kl+m)x+(lk^2 +mk+n)=0 ⇒x^2 +(2k+(m/l))x+(k^2 +((mk+n)/l))=0 ⇒(b/a)=2k+(m/l) ⇒k=(1/2)((b/a)−(m/4))
$${ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}\Leftrightarrow{x}=\alpha,\beta \\ $$$$\Rightarrow{x}^{\mathrm{2}} +\frac{{b}}{{a}}{x}+\frac{{c}}{{a}}=\mathrm{0}\Leftrightarrow{x}=\alpha,\beta \\ $$$${lx}^{\mathrm{2}} +{mx}+{n}=\mathrm{0}\:\Leftrightarrow\:{x}=\alpha+{k},\beta+{k} \\ $$$$\Rightarrow{l}\left({x}+{k}\right)^{\mathrm{2}} +{m}\left({x}+{k}\right)+{n}=\mathrm{0}\Leftrightarrow{x}=\alpha,\beta \\ $$$$\Rightarrow{l}\left({x}^{\mathrm{2}} +\mathrm{2}{xk}+{k}^{\mathrm{2}} \right)+{mx}+{mk}+{n}=\mathrm{0} \\ $$$${lx}^{\mathrm{2}} +\left(\mathrm{2}{kl}+{m}\right){x}+\left({lk}^{\mathrm{2}} +{mk}+{n}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +\left(\mathrm{2}{k}+\frac{{m}}{{l}}\right){x}+\left({k}^{\mathrm{2}} +\frac{{mk}+{n}}{{l}}\right)=\mathrm{0} \\ $$$$\Rightarrow\frac{{b}}{{a}}=\mathrm{2}{k}+\frac{{m}}{{l}} \\ $$$$\Rightarrow{k}=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{b}}{{a}}−\frac{{m}}{\mathrm{4}}\right) \\ $$$$ \\ $$
Answered by cortano12 last updated on 21/Dec/23
{ ((ax^2 +bx+c=0 ⇒α+β=−(b/a))),((lx^2 +mx+n=0⇒α+β+2k=−(m/l))) :} ⇒−(b/a) + 2k =−(m/l) ⇒2k = (b/a)−(m/l) ⇒ k =(1/2) ((b/a) − (m/l) )
$$\:\:\begin{cases}{{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}\:\Rightarrow\alpha+\beta=−\frac{{b}}{{a}}}\\{{lx}^{\mathrm{2}} +{mx}+{n}=\mathrm{0}\Rightarrow\alpha+\beta+\mathrm{2}{k}=−\frac{{m}}{{l}}}\end{cases} \\ $$$$\:\Rightarrow−\frac{{b}}{{a}}\:+\:\mathrm{2}{k}\:=−\frac{{m}}{{l}} \\ $$$$\:\Rightarrow\mathrm{2}{k}\:=\:\frac{{b}}{{a}}−\frac{{m}}{{l}} \\ $$$$\:\Rightarrow\:{k}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\left(\frac{{b}}{{a}}\:−\:\frac{{m}}{{l}}\:\right) \\ $$

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