## Monday, April 25, 2022

A new exam to cover a topic "System of General Inequalities" has been released to UNIZOR.COM - Math 4 Teens - Algebra. It contains six multiple choice problems. Highly recommended for those who seek Knowledge.

1. What is a solution to the following inequality?
(x2 + y2 −16)·(x2 + y2 − 36) < 0

2. What is the area S of the part of a coordinate plane defined by the following inequality?
(x2 + y2 − 25)·(x2 + y2 − 49) > 0

3. What is the area S of the part of a coordinate plane defined by the following system of inequalities?
x2 + y2 − 4 < 0
y − x < 0

4. What is the perimeter P of the part of a coordinate plane defined by the following system of inequalities?
x2 − 4 < 0
y2 − 16 < 0

5. What is the perimeter P of the part of a coordinate plane defined by the following system of inequalities?
x2 + y2 − 9 < 0
x + y < 0

6. The part of a coordinate plane is defined by the following inequalities?
x2 + y2 − R2 < 0
y − |x| < 0

What is the value of parameter R, if the area of thus defined part of a coordinated plane equals to 48π?

## Sunday, April 24, 2022

### Sequence Problem

PROBLEM
Let
X0 = 0,
X1 = 1
and general formula for this sequence is
Xn+2 = 3·Xn+1 − 2·Xn
Prove that
Yn = Xn2 + 2(n+2) are exact squares of some natural numbers.
PROOF
Let's prove by induction that Xn = 2n − 1
(a) For n=0 this is true since x0 = 20 − 1 = 0, as set above
(b) For n=1 this is true since x1 = 21 − 1 = 1, as set above
(c) Assume that Xk = 2k − 1 and Xk+1 = 2k+1 − 1
and, based on that, calculate Xk+2:
Xk+2 = 3·Xk+1 − 2·Xk =
= 3·(2k+1 − 1) − 2·(2k − 1) =
= 3·2k+1 − 3 − 2·2k + 2 =
= 3·2k+1 − 2k+1 − 1 =
= 2·2k+1 − 1 =
= 2k+2 − 1

which corresponds to a formula that we are trying to prove.

Using this general formula for Xk = 2k − 1, we will prove that
Yn = Xn2 + 2(n+2) are exact squares of some natural numbers.
Indeed,
Yn = (2n − 1)2 + 2(n+2) =
= 22n − 2·2n + 1 + 2(n+2) =
= 22n − 2·2n + 1 + 4·2n =
= 22n + 2·2n + 1 =
= (2n + 1)2

### Exponential equation

4x + 6x = 9x
(2·2)x + (2·3)x = (3·3)x
Since (2·3)x never equals to zero, we can divide both sides of this equation by it without losing any solutions.
The result is
(2/3)x + 1 = (3/2)x
Let y=(2/3)x
Then y + 1 = 1/y
Since y is always positive, we can multiply both sides by it getting an equivalent equation
y² + y − 1 = 0
Solutions to this quadratic equation are
y1,2 = (−1 ± √5)/2
Therefore, x1,2 = log2/3[(−1 ± √5)/2]

### Simple problem on an area of a triangle

Add two altitudes towards the bases of both triangles (big and small), one H from the top of the big triangle onto a side of 8+4=12 length and another h from the top of the small triangle onto a side of 8 length.
Let the unknown area of a small triangle be x.
Then
(a) 12·H = x + 36
(b) 8·h = x
(c) H/h = (8+3)/3
This is a simple system of three equations with three unknowns.
Divide the first equation (a) by the second (b):
(12·H)/(8·h) = (x + 36)/x
or
(d) (3/2)·(H/h) = 1 + 36/x
Using the third equation (c) that states that H/h = 11/3, express (d) as
(3/2)·(11/3) = 1 + 36/x
or
11/2 = 1 + 36/x
or
9/2 = 36/x
from which
x = 8