Algebra+ 04: UNIZOR.COM - Math+ & Problems - Algebra

Notes to a video lecture on http://www.unizor.com

Algebra+ 04

Problem A

Prove that
if X+Y+Z=1
then X²+Y²+Z² ≥ 1/3.

Hint A1
X² + Y² ≥ 2·X·Y.

Hint A2
X²+Y²+Z² =
= (X−a)²+(Y−a)²+(Z−a)²+
+2·a·(X+Y+Z)−3a² ≥
≥ 2a −3a²
Quadratic polynomial 2a−3a² has roots a=0 and a=2/3 and maximum at a=1/3 with value 2·(1/3)−3·(1/3)²=1/3.


Problem B

Solve an equation
X⁸ + X⁴ + 1 = 0

Hint B1
Substitute y=X⁴.

Hint B2
Represent the left part as a product of 4 polynomials of a second degree by adding and subtracting X⁴.

Answer B
X_1,2,3,4 = ±1/2 ± i·√3/2
X_5,6,7,8 = ±√3/2 ± i/2


Problem C

Find a number from 10 to 99 knowing that its square equals to sum of its digits in cube.

Hint C
Actually, we have to solve the equation
(10·X+Y)² = (X+Y)³
for natural
1 ≤ X ≤ 9 and 0 ≤ Y≤ 9.

Answer C
The only solution is number 27.
Check:
27² = 729
(2+7)³ = 729