Question

Stiind ca x1=x2, determinati parametrul real m:
(m+2)x^2 - 12x + (4m+1) =0

Answer 1

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Answer 2

$\it (m+2)x^2-12x+4m+1=0$

Din relațiile lui Viète, rezultă:

[tex]\it x_1+x_2=-\dfrac{b}{a} = -\dfrac{-12}{m+2} =\dfrac{12}{m+2} \ \ \ \ (1) \\\;\\ \\\;\\ Dar\ \ x_1=x_2 = t \Rightarrow x_1+x_2=2t \ \ \ \ (2) \\\;\\ \\\;\\ (1),\ (2) \Rightarrow 2t=\dfrac{12}{m+2} |_{:2} \Rightarrow t=\dfrac{6}{m+2}\ \ \ (3)[/tex]

Știm că t, din relația (3), este soluție (dublă) a ecuației date, deci:

[tex]\it (m+2)\cdot\dfrac{36}{(m+2)^2} -\dfrac{12\cdot6}{m+2}+4m+1=0|_{\cdot(m+2)} \Rightarrow \\\;\\ \\\;\\ \Rightarrow 36-72+(4m+1)(m+2) = 0 \ \Rightarrow 4m^2+9m-34=0 \Rightarrow \\\;\\ \\\;\\ \Rightarrow 4m^2-8m+17m-34=0 \Rightarrow 4(m-2) + 17(m-2) =0 \Rightarrow[/tex]


[tex]\it \Rightarrow (m-2)(4m+17) =0 \Rightarrow \begin{cases} \it m-2=0 \Rightarrow m=2 \\\;\\ \it 4m+17 =0 \Rightarrow m=-\dfrac{17}{4}\end{cases} \Rightarrow \\\;\\ \\\;\\ \Rightarrow m \in \left\{-\dfrac{17}{4},\ \ 2 \right\}[/tex]