Radical equations and inequalities

9000023704

Level: 
A
Identify a true statement which concerns the following equation. \[ \sqrt{x + 20} = 4 \]
The solution is from the set \(B = \left \{x\in \mathbb{R} : -6\leq x\leq - 2\right \}\).
The solution is from the set \(A = \left \{x\in \mathbb{R} : -4 < x\leq - 1\right \}\).
The solution is from the set \(C = \left \{x\in \mathbb{R} : -7\leq x\leq - 5\right \}\).
The solution is from the set \(D = \left \{x\in \mathbb{R} : -3 < x < 0\right \}\).

9000020006

Level: 
A
Identify a true statement referring to the solution of the following equation. \[ \sqrt{3x - 8} = x - 6 \]
The equation has a unique solution and this solution is an odd number.
The equation has two solutions, the sum of these solutions is divisible by \(5\).
The equation has a unique solution and this solution is an even number.
The equation does not have a solution in \(\mathbb{R}\).

9000020007

Level: 
A
Identify a true statement referring to the solution of the following equation. \[ \sqrt{x^{2 } - 4} = x + 1 \]
The equation does not have a solution in \(\mathbb{R}\).
The equation has a unique negative solution.
The equation has a unique positive solution.
The equation has two solutions.

9000020008

Level: 
A
Identify a true statement referring to the solution of the following equation. \[ 6x - 13\sqrt{x} + 6 = 0 \] Hint: Use the substitution \(y = \sqrt{x}\).
The solutions \(x_{1}\) and \(x_{2}\) satisfy \(x_{1} = \frac{1} {x_{2}} \).
The equation has a unique solution \(x_{1}\). This solution satisfies \(x_{1} < 1\).
The equation has a unique solution \(x_{1}\). This solution satisfies \(x_{1} > 1\).
The equation does not have a solution in \(\mathbb{R}\).