Ta có:
`2x^2+7x+11`
`=2(x^2+7/2x+11/2)`
`=2[(x^2+2*x*7/4+49/16)+39/16]`
`=2(x+7/4)^2+39/8`
Vì: `(x+7/4)^2>=0` với mọi x
`=>2(x+7/4)^2>=0` với mọi x
`=>2(x+7/4)^2+39/8>=39/8>0` với mọi x
`=>` Vô nghiệm
\(2x^2+7x+11=0\)
\(\Leftrightarrow2\left(x^2+\dfrac{7}{2}x+\dfrac{11}{2}\right)=0\)
\(\Leftrightarrow x^2+\dfrac{7}{2}x+\dfrac{11}{2}=0\)
\(\Leftrightarrow x^2+2.x.\dfrac{7}{4}+\left(\dfrac{7}{4}\right)^2+\dfrac{39}{16}=0\)
\(\Leftrightarrow\left(x+\dfrac{7}{4}\right)^2+\dfrac{39}{16}=0\)
\(\Leftrightarrow\left(x+\dfrac{7}{4}\right)^2=-\dfrac{39}{16}\)
\(\Rightarrow\)Vô lý vì \(\left(x+\dfrac{7}{4}\right)^2\ge0\forall x\)
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