Question

tìm nghiệm của đa thức 4x^2+6x-1

Answer 1

\(4x^2+6x-1=\left(2x\right)^2+2.2x.\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{13}{4}\)

\(\Leftrightarrow\left(2x+\dfrac{3}{2}\right)^2-\dfrac{13}{4}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{3}{2}=\sqrt{\dfrac{13}{4}}\\2x+\dfrac{3}{2}=-\sqrt{\dfrac{13}{4}}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3+\sqrt{13}}{4}\\x=\dfrac{-3-\sqrt{13}}{4}\end{matrix}\right.\)

Answer 2

4x² + 6x - 1 = 0

△^'= 3² - ( -1).4 = 13

vì △^' > 0 nên pt có 2 nghiệm phân biệt:

⇒\(\left\{{}\begin{matrix}x_1=\dfrac{-3+\sqrt{13}}{4}\\x_2=\dfrac{-3-\sqrt{13}}{4}\end{matrix}\right.\)

vậy,.....