Question

[Sách Giáo Khoa]

Xét dấu các tam thức bậc hai sau :

a) \(2x^2+5x+2\)

b) \(4x^2-3x-1\)

c) \(-3x^2+5x+1\)

d) \(3x^2+x+5\)

Answer 1

[ngonhuminh]

a)

\(A=2x^2+5x+2\) \(\Delta=25-16=9\)

Nếu \(\left[{}\begin{matrix}x=-2\\x=\dfrac{-1}{4}\end{matrix}\right.\) \(\Rightarrow A=0\)

nếu \(\left[{}\begin{matrix}x< -2\\x>-\dfrac{1}{4}\end{matrix}\right.\)\(\Rightarrow A>0\)

Nếu \(-2< x< \dfrac{1}{4}\Rightarrow A< 0\)

Answer 2

[ngonhuminh]

b) \(B=4x^2-3x-1\) {a+b+c=0}

Nếu x={1,-1/4} => B=0

Nếu x<-1/4 hoặc x>1 thì B>0

Nếu -1/4<x<1 thì B<0

c)

\(C=-3x^2+5x+1\) \(\Delta=25+12=37\)

\(\left[{}\begin{matrix}x=\dfrac{5+\sqrt{37}}{6}\\x=\dfrac{5-\sqrt{37}}{6}\end{matrix}\right.\) \(\Rightarrow C=0\)

\(\dfrac{5-\sqrt{37}}{6}< x< \dfrac{5+\sqrt{37}}{6}\Rightarrow C>0\)

\(\left[{}\begin{matrix}x>\dfrac{5+\sqrt{37}}{6}\\x< \dfrac{5-\sqrt{37}}{6}\end{matrix}\right.\) \(\Rightarrow C< 0\)