设方程3x2-x-5=0的两根为x₁,x₂,求x₁2+x₂2,(x₁-2)(x₂-2)
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/29 02:14:17
![设方程3x2-x-5=0的两根为x₁,x₂,求x₁2+x₂2,(x₁-2)(x₂-2)](/uploads/image/z/12619869-69-9.jpg?t=%E8%AE%BE%E6%96%B9%E7%A8%8B3x2-x-5%3D0%E7%9A%84%E4%B8%A4%E6%A0%B9%E4%B8%BAx%26%238321%3B%2Cx%26%238322%3B%2C%E6%B1%82x%26%238321%3B2%2Bx%26%238322%3B2%2C%28x%26%238321%3B-2%29%28x%26%238322%3B-2%29)
设方程3x2-x-5=0的两根为x₁,x₂,求x₁2+x₂2,(x₁-2)(x₂-2)
设方程3x2-x-5=0的两根为x₁,x₂,求x₁2+x₂2,(x₁-2)(x₂-2)
设方程3x2-x-5=0的两根为x₁,x₂,求x₁2+x₂2,(x₁-2)(x₂-2)
利用韦达定理。两根之和x1+x2=1/3,两根之积为-5/3
(x1)²+(x2)²=(x1+x2)²-2(x1)(x2)=(1/3)²-2×(-5/3)=1/9+10/3=31/9
(x1-2)(x2-2)=x1x2-2(x1+x2)+4=-5/3-2×1/3+4=4-7/3=5/3