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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
![(x+\sqrt[3]{5})(x^2-\sqrt[3]{5}x+\sqrt[3]{25})(x+1)(x^2-x+1)=0](https://tex.z-dn.net/?f=%20%28x%2B%5Csqrt%5B3%5D%7B5%7D%29%28x%5E2-%5Csqrt%5B3%5D%7B5%7Dx%2B%5Csqrt%5B3%5D%7B25%7D%29%28x%2B1%29%28x%5E2-x%2B1%29%3D0%20%20%20)
.
Here ![x_1=-\sqrt[3]{5} , x_2=-1](https://tex.z-dn.net/?f=%20x_1%3D-%5Csqrt%5B3%5D%7B5%7D%20%2C%20x_2%3D-1%20)
and you can sheck whether quadratic trinomials have real roots:
1. ![D_1=(-\sqrt[3]{5}) ^2-4\cdot \sqrt[3]{25}=\sqrt[3]{25} -4\sqrt[3]{25} =-3\sqrt[3]{25}](https://tex.z-dn.net/?f=%20D_1%3D%28-%5Csqrt%5B3%5D%7B5%7D%29%20%5E2-4%5Ccdot%20%5Csqrt%5B3%5D%7B25%7D%3D%5Csqrt%5B3%5D%7B25%7D%20-4%5Csqrt%5B3%5D%7B25%7D%20%3D-3%5Csqrt%5B3%5D%7B25%7D%20%20%3C0%20)
.
2. 
.
This means that quadratic trinomials don't have real roots.
Answer:
If you need complex roots, then
![x_{3,4}=\dfrac{\sqrt[3]{5}\pm i\sqrt{3\sqrt[3]{25}}}{2} ,\\ \\x_{5,6}=\dfrac{1\pm i\sqrt{3}}{2}](https://tex.z-dn.net/?f=%20x_%7B3%2C4%7D%3D%5Cdfrac%7B%5Csqrt%5B3%5D%7B5%7D%5Cpm%20i%5Csqrt%7B3%5Csqrt%5B3%5D%7B25%7D%7D%7D%7B2%7D%20%20%20%2C%5C%5C%20%5C%5Cx_%7B5%2C6%7D%3D%5Cdfrac%7B1%5Cpm%20i%5Csqrt%7B3%7D%7D%7B2%7D%20%20%20%20)
.
Y=square root of x
All the others would be a straight line with an unchanging slope (meaning they're linear). Y=rootx is not linear.
I hope this Helps!
1 cm because 1 cm * 1 cm * 1 cm = 1 cm
Answer:
Option (2).
Step-by-step explanation:
It is given in the question,
ΔLMN is a right triangle with base LM = 3a units
Hypotenuse MN = 5a
By applying Pythagoras theorem in ΔLMN,
MN² = LM² + NM²
(5a)² = (3a)² + MN²
25a² - 9a² = MN²
MN = √16a²
MN = 4a
Therefore, vertices of the triangle will be L(0, 0), M(3a, 0) and N(0, 4a).
Option (2) will be the answer.
