Consider the equation
.
First, you can use the substitution
, then
and equation becomes
. This equation is quadratic, so
.
Then you can factor this equation:
.
Use the made substitution again:
.
You have in each brackets the expression like
that is equal to
. Thus,
![x^3+5=(x+\sqrt[3]{5})(x^2-\sqrt[3]{5}x+\sqrt[3]{25}) ,\\x^3+1=(x+1)(x^2-x+1)](https://tex.z-dn.net/?f=%20x%5E3%2B5%3D%28x%2B%5Csqrt%5B3%5D%7B5%7D%29%28x%5E2-%5Csqrt%5B3%5D%7B5%7Dx%2B%5Csqrt%5B3%5D%7B25%7D%29%20%2C%5C%5Cx%5E3%2B1%3D%28x%2B1%29%28x%5E2-x%2B1%29%20%20%20)
and the equation is
.
Here
and you can sheck whether quadratic trinomials have real roots:
1.
.
2.
.
This means that quadratic trinomials don't have real roots.
Answer:
If you need complex roots, then
.
Answer:
Y=2/3x-1
Step-by-step explanation:
Answer:

Step-by-step explanation:
let x=11.510510510...
multiply by 1000
1000x=11510.510510...
again multiply by 1000
1000,000 x=11510510.510510...
subtract
999,000 x=1,1 49 9,000
x=11,499,000/999000=3,833,000/333,000=3833/333
333 ) 3833 (11
-333
--------
503
- 333
-------
170
------
Answer:
p = (2h-W)/-3
Step-by-step explanation:
2h - W = -3p
divide both sides by -3
(2h-W)/-3 = p
Answer:
wow
Step-by-step explanation: