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:
x = -30
Step-by-step explanation:
x - (-10) = -20
Subtracting a negative is adding
x+10 = -20
Subtract 10 from each side to isolate x
x+10-10 = -20-10
x = -30
There is no solutions, I think you got the signs mixed up.
Answer:
1:8
Step-by-step explanation:
75min/10hr
10*60=600
75/600
1/8
Answer:
B) 190 cm^2
Step-by-step explanation:
Im pretty sure all you gotta do is
20x10 which is 200 then do 5x2
which is 10 and you minus 10 from 200
Please correct me if im wrong here
I dont really remember doing these