9000028307
Level:
B
Solve the following equation.
\[
x^{3} + 6x^{2} - 8x = 0
\]
\(0\),
\(- 3 -\sqrt{17}\),
\(- 3 + \sqrt{17}\)
\(0\),
\(3 -\sqrt{17}\),
\(3 + \sqrt{17}\)
\(0\),
\(- 3\),
\(\sqrt{
17}\)
\(0\),
\(3\),
\(-\sqrt{17}\)
B
9000029301
Level:
B
Find the solution set of the following inequality.
\[
\left (x - 1\right )\left (x - 2\right )\left (x - 3\right )\geq 0
\]
\(\left [ 1;2\right ] \cup \left [ 3;\infty \right )\)
\(\left (-\infty ;\infty \right )\)
\(\left (-\infty ;1\right )\cup \left (2;3\right )\)
\(\emptyset \)
\(\{0\}\)
9000028308
Level:
B
Solve the following equation.
\[
x^{4} - 20x^{2} + 99 = 0
\]
\(-\sqrt{11}\),
\(- 3\),
\(3\),
\(\sqrt{
11}\)
\(0\),
\(- 3 -\sqrt{17}\),
\(- 3 + \sqrt{17}\)
\(0\),
\(3 -\sqrt{17}\),
\(3 + \sqrt{17}\)
\(-\sqrt{17}\),
\(- 3\),
\(3\),
\(\sqrt{
17}\)
9000029302
Level:
B
Find the solution set of the following inequality.
\[
x^{4} - 16 > 0
\]
\(\mathbb{R}\setminus \left [ -2;2\right ] \)
\(\mathbb{R}\)
\(\left (-\infty ;-4\right )\cup \left (4;\infty \right )\)
\(\left (-2;2\right )\)
\(\left (-4;4\right )\)
9000028310
Level:
B
Find the sum of all real solutions of the following equation.
\[
x^{4} - 20x^{2} + 64 = 0
\]
\(0\)
\(- 10\)
\(4\)
\(10\)
9000029305
Level:
B
Find the solution set of the following inequality.
\[
x^{4} + 81\leq 0
\]
\(\emptyset \)
\(0\)
\(\mathbb{R}\setminus \left (-9;9\right )\)
\(\mathbb{R}\)
\(\left (-\infty ;-3\right ] \cup \left [ 3;\infty \right )\)
9000029303
Level:
B
In the following list identify an inequality which does not have solution in
\(\mathbb{R}\).
\(x^{4} + 81 < 0\)
\((x - 3)^{3} > 0\)
\(x^{3} - 9x < 0\)
\(4x^{4} - 64 > 0\)
9000029308
Level:
B
Find the solution set of the following inequality.
\[
x^{3} + 4x < 0
\]
\((-\infty ;0)\)
\((2;\infty )\)
\(\mathbb{R}\)
\(\emptyset \)
\((0;\infty )\)
9000026407
Level:
B
Assuming \(x\in \left [ \frac{1}
{2};\infty \right )\),
write the following equation in the form which does not contain an absolute value.
\[
|x + 1| + |2x - 1| = 3
\]
\(x + 1 + 2x - 1 = 3\)
\(x + 1 - 2x - 1 = 3\)
\(x + 1 - 2x + 1 = 3\)
\(- x - 1 + 2x - 1 = 3\)
9000026409
Level:
B
Consider the following equation.
\[
|2x - 4| = 5x - 7
\]
Solving the equation on the intervals where it is possible to evaluate the absolute
value we get equations on partial subintervals as follows.
\[\begin{aligned}
\text{for }x &\in (-\infty ;2)\colon &\text{for }x &\in [ 2;\infty )\colon & & & &
\\ - 2x + 4 & = 5x - 7 &2x - 4 & = 5x - 7 & & & &
\\ - 7x & = -11 & - 3x & = -3 & & & &
\\x & = \frac{11}
{7} &x & = 1 & & & &
\end{aligned}\]
Find the solution set of the original equation.
\(\left \{\frac{11}
{7} \right \}\)
\(\left \{\frac{11}
{7} ;1\right \}\)
\(\left \{1\right \}\)
\(\emptyset \)