Yes; for example triangles, squares, and rectangle.
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:
a) -4
b) 2
Step-by-step explanation:
a) f(-2) = 3*-2 + 2 = -4
b) f(x) = 8
3x + 2 = 8
3x = 8 -2
3x = 6
x = 6/3
x = 2
Answer:
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Step-by-step explanation:
hi I was just thinking of you ❣️☑️☑️☑️☑️☑️♥️♥️☹️☹️ jsjsjdiie of my friends has to go back in the following weekend I am also going on Didi ❣️♥️♥️✔️Emma love myself
Answer:
option 3
Step-by-step explanation:
![(4 + 3) \sqrt[5]{x {}^{2}y } = 7 \sqrt[5]{ {x}^{2}y }](https://tex.z-dn.net/?f=%284%20%2B%203%29%20%5Csqrt%5B5%5D%7Bx%20%7B%7D%5E%7B2%7Dy%20%7D%20%3D%207%20%20%5Csqrt%5B5%5D%7B%20%7Bx%7D%5E%7B2%7Dy%20%7D%20)