Question

\(2x.\left(x+\dfrac{4}{3}\right)=\dfrac{1}{6}\)

Answer 1

\(2x\left(x+\dfrac{4}{3}\right)=\dfrac{1}{6}\)

=>\(x\left(x+\dfrac{4}{3}\right)=\dfrac{1}{12}\)

=>\(x^2+\dfrac{4}{3}x-\dfrac{1}{12}=0\)

\(\text{Δ}=\left(\dfrac{4}{3}\right)^2-4\cdot1\cdot\left(-\dfrac{1}{12}\right)=\dfrac{16}{9}+3=\dfrac{43}{9}>0\)

Do đó, phương trình có hai nghiệm có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-\dfrac{4}{3}-\sqrt{\dfrac{43}{9}}}{2}=\dfrac{-4-\sqrt{43}}{6}\\x_2=\dfrac{-4+\sqrt{43}}{6}\end{matrix}\right.\)