Question

Chứng minh phương trình x²+3x+4=0 vô nghiệm

Answer 1

\(x^2+3x+4=0\)

\(\Leftrightarrow x^2+2.x.\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{7}{4}=0\)

\(\Leftrightarrow\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}=0\)

\(\Leftrightarrow\left(x+\dfrac{3}{2}\right)^2=-\dfrac{7}{4}\left(VL\right)\)

Vậy ĐPCM

Answer 2

\(x^2+3x+4=0\Leftrightarrow x^2+2.x.\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{7}{4}=0\)

\(\Leftrightarrow x^2+2.x.\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2+\dfrac{7}{4}=0\Leftrightarrow\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}=0\)

Ta có \(\left(x+\dfrac{3}{2}\right)^2\ge0,\forall x\)

\(\Rightarrow\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}>0,\forall x\)

Vậy phương trình vô nghiệm.