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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

{\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})