Hello there.
First, assume the numbers
such that they satisties both affirmations:
- The sum of the squares of two numbers is
. - The product of the two numbers is
.
With these informations, we can set the following equations:
![\begin{center}\align x^2+y^2=8\\ x\cdot y=4\\\end{center}](https://tex.z-dn.net/?f=%5Cbegin%7Bcenter%7D%5Calign%20x%5E2%2By%5E2%3D8%5C%5C%20x%5Ccdot%20y%3D4%5C%5C%5Cend%7Bcenter%7D)
Multiply both sides of the second equation by a factor of
:
![2\cdot x\cdot y = 2\cdot 4\\\\\\ 2xy=8~~~~~(2)^{\ast}](https://tex.z-dn.net/?f=2%5Ccdot%20x%5Ccdot%20y%20%3D%202%5Ccdot%204%5C%5C%5C%5C%5C%5C%202xy%3D8~~~~~%282%29%5E%7B%5Cast%7D)
Make ![(1)-(2)^{\ast}](https://tex.z-dn.net/?f=%281%29-%282%29%5E%7B%5Cast%7D)
![x^2+y^2-2xy=8-8\\\\\\ x^2-2xy+y^2=0](https://tex.z-dn.net/?f=x%5E2%2By%5E2-2xy%3D8-8%5C%5C%5C%5C%5C%5C%20x%5E2-2xy%2By%5E2%3D0)
We can rewrite the expression on the left hand side using the binomial expansion in reverse:
, such that:
![(x-y)^2=0](https://tex.z-dn.net/?f=%28x-y%29%5E2%3D0)
The square of a number is equal to
if and only if such number is equal to
, thus:
![x-y=0\\\\\\ x=y~~~~~~(3)](https://tex.z-dn.net/?f=x-y%3D0%5C%5C%5C%5C%5C%5C%20x%3Dy~~~~~~%283%29)
Substituting that information from
in
, we get:
![x\cdot x = 4\\\\\\ x^2=4](https://tex.z-dn.net/?f=x%5Ccdot%20x%20%3D%204%5C%5C%5C%5C%5C%5C%20x%5E2%3D4)
Calculate the square root on both sides of the equation:
![\sqrt{x^2}=\sqrt{4}\\\\\\ |x|=2\\\\\\ x=\pm~2](https://tex.z-dn.net/?f=%5Csqrt%7Bx%5E2%7D%3D%5Csqrt%7B4%7D%5C%5C%5C%5C%5C%5C%20%7Cx%7C%3D2%5C%5C%5C%5C%5C%5C%20x%3D%5Cpm~2)
Once again with the information in
, we have that:
![y=\pm~2](https://tex.z-dn.net/?f=y%3D%5Cpm~2)
The set of solutions of that satisfies both affirmations is:
![S=\{(x,~y)\in\mathbb{R}^2~|~(x,~y)=(-2,\,-2),~(2,~2)\}](https://tex.z-dn.net/?f=S%3D%5C%7B%28x%2C~y%29%5Cin%5Cmathbb%7BR%7D%5E2~%7C~%28x%2C~y%29%3D%28-2%2C%5C%2C-2%29%2C~%282%2C~2%29%5C%7D)
This is the set we were looking for.
Answer:
b=(13/4, -27/4, -4)
Step-by-step explanation:
A.x=b
x=(a, b, c)
a - b + 2c = 2
3a + b + 0 = 3
0 + 0 + c = -4
solving this system
c=-4
a-b=10, 3a+b=3
a=10+b, 3(10+b)+b=3
b=-27/4
a=10-27/4
a=13/4
Answer: 5
Step-by-step explanation:
M= 5
Answer:
D
Step-by-step explanation: