[tex]57 d) \ \ \sqrt{2x^2+x+1}=2x^2+x-1
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Notam:\ 2x^2+x+1 =t, \ t\ \textgreater \ 0
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2x^2+x-1=2x^2+x+1-2=t-2,\ t-2\ \textgreater \ 0\ \Rightarrow t\ \textgreater \ 2
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Ecuatia\ devine:
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\sqrt{t} =t-2 \Rightarrow (\sqrt{t})^2 =(t-2)^2 \Rightarrow t=t^2-4t+4 \Rightarrow t^2-5t+4=0[/tex]
[tex]\Rightarrow t_1=1,\ nu\ convine\ (pentru\ ca\ t\ \textgreater \ 2)
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t_2=4
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Revenim\ asupra\ notatiei:
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t=4 \Rightarrow 2x^2+x+1=4 \Rightarrow 2x^2+x-3=0
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Obtinem:
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\it x_1=-\dfrac{3}{2},\ \ x_2=1.
[/tex]
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