a*****g
发帖数: 1320 | 1 1) The solutions of x2 +bx+c=0 are each 5 more than the solution of x2+7x+3=
0. What are the value of b and c? Express your answer as an ordered pair (b,
c)
2) A cubic equation of the form x3+ bx2+cx+d==0 has solutions x=3, x=4 and x
=5. What are the value of b, c, and d? Express your answer as an ordered
triple (b,c,d)
3) What is the sum of the reciprocals of the solutions of x3-3x2-13x+15=0?
Express your answer as a common fraction
4) What is the sum of the squares of the solutions of x3-15x2+66x-80=0?
5) The solutions of X3-63x2+cx-1728=0 form a geometric sequence. What is
thevalue of c?
Thanks a lot! | m*********y
发帖数: 1735 | 2 1)
假设第二个方程的解是x1,x2，所以第一个方程的解就是(x1+5),(x2+5)
所以，x1+x2=-7,x1*x2=3;
b=-(x1+5)+(x2+5)=-(x1+x2)-10=-3
c=(x1+5)*(x2+5)=x1*x2+5(x1+x2)+25=3-35+25=-7
2)
{27+9b+3c+d=0 (1)
64+16b+4c+d=0;(2)
125+25b+5c+d=0;(3)
}
(2)-(1)=(4)
(3)-(2)=(5)
(5)-(4)=(6)
倒回去
b=-12 c=47; d=-60;
3) 假设三个根是x1,x2,x3
(x-x1)(x-x2)(x-x3)=0
x1x2+x2x3+x1x3=-13
x1+x2+x3=3;
x1x2x3=-15;
1/x1+1/x2+1/x3=(x1x2+x2x3+x1x3)/(x1x2x3)=13/15
4)the same way as above
x1^2+x2^2+x3^2=(x1+x2+x3)^2-(x1x2+x2x3+x1x3)=15^2-66=159
5)
almost the same
first solution is x
second is ax
so the thrid one is a^2x
then x*ax*a^2x=(ax)^3=1728
ax=12
so x+ax+a^2x=63
so a=1/4
so, three solutions are 3,12,48
so, c=36+144+12*48=756
结果不见得正确，你让你孩子再重新验算一遍。。。
3=
b,
x
【在 a*****g 的大作中提到】
: 1) The solutions of x2 +bx+c=0 are each 5 more than the solution of x2+7x+3=
: 0. What are the value of b and c? Express your answer as an ordered pair (b,
: c)
: 2) A cubic equation of the form x3+ bx2+cx+d==0 has solutions x=3, x=4 and x
: =5. What are the value of b, c, and d? Express your answer as an ordered
: triple (b,c,d)
: 3) What is the sum of the reciprocals of the solutions of x3-3x2-13x+15=0?
: Express your answer as a common fraction
: 4) What is the sum of the squares of the solutions of x3-15x2+66x-80=0?
: 5) The solutions of X3-63x2+cx-1728=0 form a geometric sequence. What is
| m*********y
发帖数: 1735 | 3 最后一个a写错了
【在 m*********y 的大作中提到】
: 1)
: 假设第二个方程的解是x1,x2，所以第一个方程的解就是(x1+5),(x2+5)
: 所以，x1+x2=-7,x1*x2=3;
: b=-(x1+5)+(x2+5)=-(x1+x2)-10=-3
: c=(x1+5)*(x2+5)=x1*x2+5(x1+x2)+25=3-35+25=-7
: 2)
: {27+9b+3c+d=0 (1)
: 64+16b+4c+d=0;(2)
: 125+25b+5c+d=0;(3)
: }
| a*****g
发帖数: 1320 | 4 Super many thanks!
【在 m*********y 的大作中提到】
: 1)
: 假设第二个方程的解是x1,x2，所以第一个方程的解就是(x1+5),(x2+5)
: 所以，x1+x2=-7,x1*x2=3;
: b=-(x1+5)+(x2+5)=-(x1+x2)-10=-3
: c=(x1+5)*(x2+5)=x1*x2+5(x1+x2)+25=3-35+25=-7
: 2)
: {27+9b+3c+d=0 (1)
: 64+16b+4c+d=0;(2)
: 125+25b+5c+d=0;(3)
: }
|
|
