Exercise 1

Solve:

\frac {2}{3} [x - (1 - \frac {x - 2}{3})] + 1 \leq x

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Exercise 2

Solve:

4x^2 - 4x + 1 \leq 0

Exercise 3

Solve:

\frac {x ^2 + 4} {x^2 - 4} \geq 0

Exercise 4

Calculate the values of k for which the roots of the equation x² − 6x + k = 0 are two real and distinct numbers.

Exercise 5

Solve:

\left\{\begin{matrix} x \geq 4\\ y \geq 2 \end{matrix}\right

 

2 \left\{\begin{matrix} x + y \geq 0 \\ 2x - y \geq 0 \end{matrix}\right

 

\left\{\begin{matrix}x + y \geq 0\\ 2x - y  \geq 0\\ x \leq 6 \end{matrix}\right

Exercise 6

Solve:

\left\{\begin{matrix} (x + 1) \cdot 10 + x \leq 6 (2x + 1) \\ 4 (x - 10) < -6(2 - x) - 6x \end{matrix}\right

 

Exercise 7

Solve:

x^4 + 12 x^3 - 64 x^2 > 0

x ^4 - 25x^2 - 144 < 0

x^4 - 16x^2 - 225 \geq 0

 

Solution of exercise 1

Solve:

\frac {2}{3} [x - (1 - \frac {x - 2}{3})] + 1 \leq x

\frac {2}{3} [x - 1 + \frac {x - 2}{3}] + 1 \leq x

\frac {2}{3}x - \frac {2}{3} + \frac {2x - 4}{9} + 1 \leq x

6x - 6 + 2x - 4 + 9 \leq 9x

-x \leq 1

x \geq -1

[-1, \infty)

 

Solution of exercise 2

Solve:

4x^2 - 4x + 1 \leq 0

4x^2 - 4x + 1 = 0

4x² − 4x + 1 ≤ 0

4x² − 4x + 1 = 0

x = \frac {4 \pm \sqrt {16 - 16}} {8} = \frac {4}{8} = \frac {1}{2}

(x - \frac {1}{2})^2 \leq 0

x = \frac {1}{2}

 

Solution of exercise 3

Solve:

\frac {x ^2 + 4} {x^2 - 4} \geq 0

 

x^2 + 4 = 0

x = \pm \sqrt {-4} \notin R

    The numerator is always positive.

x^2 - 4 = 0

x = \pm \sqrt{4}

x_1 = 2

x_2 = -2

The denominator cannot be zero.

\frac{1}{x^2 - 4} \geq 0

x \neq -2

x \neq 2

Therefore, the original inequality will be equivalent to:

x^2 - 4 > 0

(- \infty, -2) U (2, +\infty)

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Solution of exercise 4

Calculate the values of k for which the roots of the equation x^2 - 6x + k = 0  are two real and distinct numbers.

(-6)^2 - 4k > 0

36 - 4k > 0

-4k > -36

k < 9

(- \infty, 9)

 

Solution of exercise 5

Solve:

\left\{\begin{matrix} x \geq 4\\ y \geq 2 \end{matrix}\right

x = 4

y = 2

2 \left\{\begin{matrix} x + y \geq 0 \\ 2x - y \geq 0 \end{matrix}\right

x + y = 0

(0, 0)

(1, - 1)

2 + 2 \geq 0

2x - y = 0

(0, 0)

(1, 2)

2 \cdot 2 - 2 \geq 0

\left\{\begin{matrix}x + y \geq 0\\ 2x - y  \geq 0\\ x \leq 6 \end{matrix}\right

x + y = 0

(0, 0)        (1, -1)

2 + 2 \geq 0

2x - y = 0

(0, 0)        (1, 2)

2 \cdot 2 - 2 \geq 0

2 \leq 6

     

 

Solution of exercise 6

Solve:

\left\{\begin{matrix} (x + 1) \cdot 10 + x \leq 6 (2x + 1) \\ 4 (x - 10) < -6(2 - x) - 6x \end{matrix}\right

(x + 1) \cdot 10 + x \leq 6 (2x + 1)

10x + 10 + x \leq 12x + 6

10x + x - 12x \leq 6 - 10

-x \leq -4         x \geq 4

4 (x - 10) < -6 (2 - x) - 6x

4x - 40 < -12 + 6x - 6x

4x - 6x + 6x < -12 + 40

4x < 28

x< 7

[4, 7)

 

Solution of exercise 7

Solve:

x^4 + 12 x^3 - 64 x^2 > 0

x^2 (x^2 + 12x - 64) > 0

As the first factor is always positive, consider the sign of the 2nd factor.

x^2 + 12x - 64 = 0

x = \frac {-12 \pm \sqrt {144 + 256}} {2}

= \frac {-12 \pm 20}{2}

x _2 = 4

x_3 = -16

P(-17) = (-17)^2 + 12 \cdot 17 - 64 > 0

P(0) = 0^2 + 12 \cdot 0 - 64 < 0

P(5) = 5^2 + 12 \cdot 5 - 64 > 0

(- \infty , -16] U [4, \infty)

 

x ^4 - 25x^2 - 144 < 0

x^4 - 25x^2 - 144 = 0

x^2 = t

t^2 - 25t + 144 = 0

t = \frac {25 \pm \sqrt {625 - 576}} {2}

= \frac {25 \pm 7}{2}

t_1 = 16

t_2 = 9

x^2 = 16

x = \pm \sqrt{16}

x_1 = 4

x_2 = -4

 

x^2 = 9

x = \pm \sqrt{9}

x_1 = 3

x_2 = -3

(-4, -3) U (3, 4)

x^4 - 16x^2 - 225 \geq 0

x^4 - 16x^2 - 225 = 0

x^2 = t

t^2 - 16t - 225 = 0

t = \frac {16 \pm \sqrt {256 + 900}} {2}

= \fracc {16 \pm 34}{2}

t_1 = 25

t_2 = -9

x^2 = 25

x = \pm \sqrt{25}

x_1 = 5

x_2 = -5

x^2 = -9

x = \pm \sqrt {-9} \notin R

(x ^2 - 25) \cdot (x^2 + 9) \geq 0

The second factor is always positive and nonzero, therefor, only consider the sign of the 1st factor.

(x^2 - 25) \geq 0

(-\infty, -5] U [5, +\infty)

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Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.