If x-y = 8 and xy=345, what are the values of x and y?
回答 (4)
2017-06-10 9:07 am
Let z be the average of x and y. Then you have
.. (z +4)(z -4) = 345
.. z^2 -4^2 = 345
.. z^2 = 345 +16 = 361
.. z = 19 . . . . from your memory of squares of numbers less than 20
x = z+4 = 23
y = z-4 = 15
_____
You can actually do all this math in your head, if you want. It's that simple.
Of course, the second solution is x=-15, y = -23, from z=-√361.
.. (z +4)(z -4) = 345
.. z^2 -4^2 = 345
.. z^2 = 345 +16 = 361
.. z = 19 . . . . from your memory of squares of numbers less than 20
x = z+4 = 23
y = z-4 = 15
_____
You can actually do all this math in your head, if you want. It's that simple.
Of course, the second solution is x=-15, y = -23, from z=-√361.
2017-06-10 7:22 am
y^2+8y-345 = 0
(y-15)(y+23) = 0
y = -23, x = -15
y = 15. x = 23
(y-15)(y+23) = 0
y = -23, x = -15
y = 15. x = 23
2017-06-10 6:44 pm
x - y = 8
x = y+8
y(y+8) = 345
y^2 + 8y - 345 = 0
(y + 4)^2 = 345 + 16 = 361
y + 4 = ±19
y = -4 ± 19
y = 15 and x = 23
or y = -23 and x = -15
x = y+8
y(y+8) = 345
y^2 + 8y - 345 = 0
(y + 4)^2 = 345 + 16 = 361
y + 4 = ±19
y = -4 ± 19
y = 15 and x = 23
or y = -23 and x = -15
2017-06-10 9:18 am
x - y = 8
xy = 345
(x - y)^2 = 8^2
(x + y)^2
= (x - y)^2 + 4xy
= 64 + 1380
x + y = (1444)^(1/2) = 38
Solution:
x = 23, y = 15
xy = 345
(x - y)^2 = 8^2
(x + y)^2
= (x - y)^2 + 4xy
= 64 + 1380
x + y = (1444)^(1/2) = 38
Solution:
x = 23, y = 15
收錄日期: 2021-04-24 00:30:05
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20170609231908AAU0Hlh