Question

一元二次方程解答题

不解方程，作一个一元二次方程，使它的两根: (1)分别是x²－2x－3＝0的两根立方（2）分别是方程3x²＋5x－10＝0的两根的3倍（3）分别是方程5x²＋2x－3＝0的两根平方的负倒数

Answer 1

（1）因为x1+x2=2,x1*x2=-3
所以x1^3+x2^3=(x1+x2)(x1^2-x1*x2+x2^2)=-3*[(x1+x2)^2-3x1*x2]
=-3*[4-3*(-3)]
=-39
x1^3*x2^3=(x1*x2)^3=(-3)^3=-27
所以方程为y^2+39y-37=0
(2)因为x1+x2=-5/3,x1*x2=-10/3
所以3x1+3x2=3*(x1+x2)=3*-5/3=-5
3x1*3x2=9x1*x2=9*(-10/3=-30
所以方程为y^2+5y-30=0
(3)因为x1+x2=-2/5,x1*x2=-3/5
所以-1/x1^2+(-1/x2^2)=-(x2^2+x1^2)/x1^2*x2^2=-[(x1+x2)^2-2x1x2]/(x1*x2)^2
=-[(-2/5)^2-2*(-3/5)]/(-3/5)^2=-34/9
-1/x1*(-1/x2)=1/x1*x2=1/(-3/5)=5/3
所以方程为y^2+34/9y+5/3=0
即9y^2+34y+15=0

Answer 2

x^2-2x-3=0
x1+x2=2
x1x2=-3
x1`=x1^3
x2`=x2^3
x1`+x2`=x1^3+x2^3=(x1+x2)(x1^2-x1x2+x2^2)=(x1+x2)((x1+x2)^2-2x1x2-x1x2)
=(x1+x2)((x1+x2)^2-3x1x2)
=2(2^2-3*(-3))=2(4+9)=2*13=26
x1`x2`=x1^3*x2^3=(x1x2)^3=(-3)^3=-27
x^2-26x-27=0
(2)3x^2+5x-10=0
x1+x2=-5/3
x1x2=-10/3
x1`=3x1,x2`=3x2
x1`+x2`=3x1+3x2=3(x1+x2)=3*(-5/3)=-5
x1`x2`=3x1*3x2=9x1x2=9*(-10/3)=-30
x^2+5x-30=0
(3)5x^2+2x-3=0
x1+x2=-2/5
x1x2=-3/5
x1`=-1/x1^2,x2`=-1/x2^2
x1`+x2`=-1/x1^2-1/x2^2=-(1/x1^2+1/x2^2)=-(x1^2+x2^2)/x1^2x2^2
=-((x1+x2)^2-2x1x2)/(x1x2)^2
=-(4/25-2*(-3/5))/9/25
=-(4/25+6/5)/9/25=-(34/25)/9/25=-34/9
x1`x2`=-1/x1^2*(-1/x2^2)=1/x1^2x2^2=1/(x1x2)^2=1/(-3/5)^2=1/9/25=25/9
x^2+34/9x+25/9=0
9x^2+34x+25=0

Answer 3

利用韦达定理
（1）设已知方程的两个根分别是x1和x2
则x1+x2=2 x1x2=-3
所求的方程的两个根分别是x1*3+x2*3
=(x1+x2)(x1*2-x1x2+x2*2)
=(x1+x2)[(x1+x2)*2-3x1x2]
=2[2*2-6]
=-4
(x1x2)*3=-27 所求方程为 x*2+4x-27=0
(2)(3)同理