1) \(4x^2-9=0\)
Theo pt ta có: \(a=4;b=0;c=-9\)
\(\Delta=b^2-4ac=0^2-4.4.\left(-9\right)=144>0\)
=> Pt có 2 nghiệm phân biệt
\(x_1=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-0-\sqrt{144}}{2.4}=-\dfrac{3}{2}\\ x_2=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-0+\sqrt{144}}{2.4}=\dfrac{3}{2}\)
2) \(-2x^2+50=0\)
Theo pt ta có: \(a=-2;b=0;c=50\)
\(\Delta b^2-4ac=0^2-4.\left(-2\right).50=400>0\)
=> PT có 2 nghiệm phân biệt
\(x_1=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-0-\sqrt{400}}{2.\left(-2\right)}=5\\ x_2=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-0+\sqrt{400}}{2a}=-5\)
3) \(3x^2+11=0\)
Theo pt ta có: \(a=3;b=0;c=11\)
\(\Delta=b^2-4ac=0^2-4.3.11=-132< 0\)
=> PT vô nghiệm
1) 4x2 - 9 = 0
=>4x2=9
=>x2=9/4
=>x=\(\pm\dfrac{3}{2}\)
2) - 2x2 + 50 = 0
=>2x2=50
=>x2=25
=>x=\(\pm5\)
3) 3x2 + 11 = 0
=>3x2=-11
=>x2=-11/3(vo li)
=>x\(\in\phi\)
1) 4x2 - 9 = 0
Δ = b2 - 4ac = 02 - 4.4.(-9) = 144 > 0
=> pt đã cho có 2 nghiệm phân biệt :
x1 = \(\dfrac{\text{ −b+√Δ}}{2a}=\dfrac{-0+\sqrt{144}}{2.4}=\dfrac{3}{2}\)
x2 =\(\dfrac{\text{ −b−√Δ}}{2a}=\dfrac{-0-\sqrt{144}}{2.4}=-\dfrac{3}{2}\)
2) - 2x2 + 50 = 0
\(\Delta=b^2-4ac\) = 02 - 4.(-2).50 = 400 > 0
=> pt có 2 nghiệm phân biệt :
x1 = \(\dfrac{-b+\sqrt{\Delta}}{2.a}=\dfrac{-0+\sqrt{400}}{2.\left(-2\right)}=-5\)
x2 = \(\text{}\text{}\dfrac{-b-\sqrt{\Delta}}{2.a}=\dfrac{-0-\sqrt{400}}{2.\left(-2\right)}=5\)
3) 3x2 + 11 = 0
Δ = b2 - 4ac = 02 - 4.3.11 = -132 < 0
=> pt vô nghiệm