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)).


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Question Number 202114 by MATHEMATICSAM last updated on 21/Dec/23

$If α and β are the roots of$ ${a x}^{2} + b x + c = 0 and α + k and β + k are$ $the roots of {l x}^{2} + m x + n = 0 then prove$ $that k = \frac{1}{2} (\frac{b}{a} - \frac{m}{l}) .$
	Answered by aleks041103 last updated on 21/Dec/23

	${a x}^{2} + b x + c = 0 \Leftrightarrow x = α, β$ $\Rightarrow x^{2} + \frac{b}{a} x + \frac{c}{a} = 0 \Leftrightarrow x = α, β$ ${l x}^{2} + m x + n = 0 \Leftrightarrow x = α + k, β + k$ $\Rightarrow l {(x + k)}^{2} + m (x + k) + n = 0 \Leftrightarrow x = α, β$ $\Rightarrow l (x^{2} + 2 x k + k^{2}) + m x + m k + n = 0$ ${l x}^{2} + (2 k l + m) x + ({l k}^{2} + m k + n) = 0$ $\Rightarrow x^{2} + (2 k + \frac{m}{l}) x + (k^{2} + \frac{m k + n}{l}) = 0$ $\Rightarrow \frac{b}{a} = 2 k + \frac{m}{l}$ $\Rightarrow k = \frac{1}{2} (\frac{b}{a} - \frac{m}{4})$
	Answered by cortano12 last updated on 21/Dec/23
	${ ((ax^2 +bx+c=0 ⇒α+β=−(b/a))),((lx^2 +mx+n=0⇒α+β+2k=−(m/l))) :} ⇒−(b/a) + 2k =−(m/l) ⇒2k = (b/a)−(m/l) ⇒ k =(1/2) ((b/a) − (m/l) )$
	${\begin{cases} {a x}^{2} + b x + c = 0 \Rightarrow α + β = - \frac{b}{a} \\ {l x}^{2} + m x + n = 0 \Rightarrow α + β + 2 k = - \frac{m}{l} \end{cases}$ $\Rightarrow - \frac{b}{a} + 2 k = - \frac{m}{l}$ $\Rightarrow 2 k = \frac{b}{a} - \frac{m}{l}$ $\Rightarrow k = \frac{1}{2} (\frac{b}{a} - \frac{m}{l})$
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