Problem 130 - College Algebra Essentials

Question

Determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. The equations \(\sqrt{x+4}=-5\) and \(x+4=25\) have the same solution set.

Step-by-Step Solution

Answer

The statement 'Find \(b\) such that \(\frac{4x-b}{x-5}=3\) has a solution set given by \(\varnothing \)' is true as is, no modifications of the statement need to be done. Any number in the place of \(b\) would still result in the denominator being zero when \(x = 5\), thereby making the equation undefined.

Step 1: Set the denominator equal to zero

The goal of this step is to find the value of \(x\) that makes the denominator of the fraction equal to zero. In this case, we'll set \(x - 5 = 0\). By solving this equation, the result will be \(x = 5\).

Step 2: Substituting \(x\) in the equation

Now, replace the value of \(x\) in the equation \(\frac{4x-b}{x-5}=3\). Substituting \(x = 5\) into the expression, we have \(\frac{4*5-b}{5-5}=3\). Simplifying the numerator gives \(20-b\) and the denominator results in \(0\). So, the equation becomes \(\frac{20-b}{0}=3\). Since division by zero is undefined in mathematics, this equation doesn't produce a solution. Thus, the statement is already true - the set of solutions for this equation is indeed the empty set.

Key Concepts

Square Root Equations

Square root equations involve the radical symbol, \(\sqrt{}\), which denotes the square root. These types of equations look like \(\sqrt{x} = a\), where \(a\) is usually a real number. Solving square root equations involves finding values of \(x\) that make the equation true.

• To solve, isolate the square root on one side of the equation.
• Then, square both sides to eliminate the square root.
• This often leads to a quadratic equation that can be solved using standard algebraic techniques.

It is important to check your solutions by substituting them back into the original equation, as squaring can sometimes introduce extraneous solutions that are not valid for the original equation. In our exercise, since \(\sqrt{x+4}=-5\) doesn't hold true (as the square root yields only non-negative numbers), it shows there are no valid solutions within the real number system.

Solution Sets

Solution sets refer to the collection of values that satisfy an equation. In algebra, determining the solution set of an equation involves finding all possible values of the variable that make the equation true.

• A solution set can be empty, finite, or infinite, depending on the equation.
• For radical equations, it's crucial to verify potential solutions since solutions fit the domain specified by the radical.

When examining the solution set of \(\sqrt{x+4}=-5\) and \(x+4=25\), we find that no run-of-the-mill, real-number solution exists for the square root equation. Thus, the solution set is empty for this statement.

Undefined Expressions

Expressions such as those involving division by zero are considered undefined in mathematics. For example, the expression \(\frac{a}{0}\) is undefined because division by zero does not yield a real or valid number.

• Identifying undefined expressions helps in recognizing when equations have no solution.
• In equations involving division, always check the denominator to ensure it is never zero.

In our step-by-step solution, we compared our expression to find \(x = 5\) produced a division by zero. Since this causes the expression \(\frac{20-b}{0}=3\) to be undefined, it confirms the solution set is indeed empty.

Equations Involving Radicals

Equations involving radicals can be tricky because they feature variables under a square root or other root. Solving these necessitates a careful approach to ensure values are valid.

• Start by isolating the radical on one side of the equation.
• Eliminate the radical by raising both sides of the equation to the same power as the root.
• After simplifying, you can solve for the variable using typical algebraic methods.

Remember to check your solutions against the original radical equation. Any time a square root is involved, it's essential to ensure no extraneous solutions remain. In our scenario, since \(\sqrt{x+4}=-5\) involves a contradiction in its structure (a square root equating to a negative), the problem highlights the importance of understanding the constraints radicals impose on equations.