Question

\(4x^2+7x+8=0\)

Tuyển ny

Answer 1

\(\left(2x\right)^2+2\cdot2x\cdot\frac{7}{4}+\frac{49}{16}=-\frac{177}{16}\)

\(\Leftrightarrow\left(2x+\frac{7}{4}\right)^2=-\frac{177}{16}\)

Pt vô nghiệm

Answer 2

\(4x^2+7x+8\)

*\(\Delta=7^2-4.4.8=49-128=-79\)

*\(\Delta< 0\)nên pt vô nghiệm