Problem 29
Question
Find all solutions of the equation algebraically. Check your solutions. $$\sqrt{x+1}=\sqrt{3 x+1}$$
Step-by-Step Solution
Verified Answer
The solution to the equation \(\sqrt{x+1} = \sqrt{3x+1}\) is \(x = 0\).
1Step 1: Set Up the Equation
Start with the equation provided: \(\sqrt{x+1} = \sqrt{3x+1}\).
2Step 2: Square Both Sides of the Equation
To get rid of the square roots, square both sides of the equation. This gives us \((\sqrt{x+1})^2 = (\sqrt{3x+1})^2\), which simplifies to \(x+1 = 3x+1\).
3Step 3: Rearrange the Equation
Subtract \(x+1\) from both sides to isolate the variable, resulting in the equation: \(2x = 0\).
4Step 4: Solve for x
Divide both sides by 2 to solve for x, which gives us: \(x = 0\).
5Step 5: Check the Solution
Substitute \(x = 0\) back into the original equation \(\sqrt{x+1} = \sqrt{3x+1}\) to check if the solution holds true. Plugging 0 into the equation results in \(\sqrt{1} = \sqrt{1}\), which is true, confirming that \(x = 0\) is the solution.
Key Concepts
Square Both SidesIsolate the VariableChecking Solutions in Algebra
Square Both Sides
When solving radical equations, such as \(\sqrt{x+1} = \sqrt{3x+1}\), one of the first steps is to 'square both sides' of the equation. This means that we take the square of both sides to eliminate the square roots. Squaring a number, or in this case a radical, reverses the process of finding a square root.
For example, in the equation above, when we square \(\sqrt{x+1}\) and \(\sqrt{3x+1}\), we get \( (\sqrt{x+1})^2 \) and \( (\sqrt{3x+1})^2 \) respectively. This results in \(x+1\) on the left side and \(3x+1\) on the right side after the square roots have been eliminated. It's crucial to remember that when you square both sides, you must apply the squaring operation to the entire side of the equation, not just the radical.
However, be cautious as this process can introduce extraneous solutions - solutions that do not actually satisfy the original equation. That is why after finding solutions by squaring both sides, we must always check each solution in the original equation.
For example, in the equation above, when we square \(\sqrt{x+1}\) and \(\sqrt{3x+1}\), we get \( (\sqrt{x+1})^2 \) and \( (\sqrt{3x+1})^2 \) respectively. This results in \(x+1\) on the left side and \(3x+1\) on the right side after the square roots have been eliminated. It's crucial to remember that when you square both sides, you must apply the squaring operation to the entire side of the equation, not just the radical.
However, be cautious as this process can introduce extraneous solutions - solutions that do not actually satisfy the original equation. That is why after finding solutions by squaring both sides, we must always check each solution in the original equation.
Isolate the Variable
Once the radicals are removed by squaring both sides, the next step in solving an equation is to isolate the variable. Isolating the variable means to manipulate the equation so that the variable of interest is on one side and everything else is on the other side.
In our example, after removing the square roots, we have been left with the equation \(x+1 = 3x+1\). To isolate \(x\), we need to get all terms with \(x\) on one side and the constants on the other. We do this by performing the same operation on both sides of the equation to maintain equality. Subtracting \(x\) from both sides, we get \(0 = 2x\), and by further subtracting \(1\) from both sides we end up with \(0 = 2x\).
This is a simple equation to solve. We divide both sides by 2, which leaves us with \(x = 0\), successfully isolating the variable. In more complex equations, you might have to perform multiple operations such as distributing, combining like terms, and using inverse operations to isolate the variable effectively.
In our example, after removing the square roots, we have been left with the equation \(x+1 = 3x+1\). To isolate \(x\), we need to get all terms with \(x\) on one side and the constants on the other. We do this by performing the same operation on both sides of the equation to maintain equality. Subtracting \(x\) from both sides, we get \(0 = 2x\), and by further subtracting \(1\) from both sides we end up with \(0 = 2x\).
This is a simple equation to solve. We divide both sides by 2, which leaves us with \(x = 0\), successfully isolating the variable. In more complex equations, you might have to perform multiple operations such as distributing, combining like terms, and using inverse operations to isolate the variable effectively.
Checking Solutions in Algebra
After solving an algebraic equation, it is imperative to verify that the potential solutions are valid. This process, known as 'checking solutions,' involves substituting the values back into the original equation to ensure that they satisfy it.
For our equation \(\sqrt{x+1} = \sqrt{3x+1}\), once we determine that \(x = 0\) is a potential solution, we check it by replacing \(x\) with 0 in the original equation: \(\sqrt{0+1} = \sqrt{3\times0+1}\). Simplifying both sides, we find that \(\sqrt{1} = \sqrt{1}\), which indicates that our solution is correct because both sides of the equation are equal.
Checking solutions defends against the possibility of extraneous solutions, which are results that emerge from the algebraic manipulations but don’t hold true in the original equation. It’s a safeguard to ensure that the solutions derived are precise and applicable. In some cases, when you find more than one solution, each must be checked separately to determine which, if any, are true solutions to the original equation.
For our equation \(\sqrt{x+1} = \sqrt{3x+1}\), once we determine that \(x = 0\) is a potential solution, we check it by replacing \(x\) with 0 in the original equation: \(\sqrt{0+1} = \sqrt{3\times0+1}\). Simplifying both sides, we find that \(\sqrt{1} = \sqrt{1}\), which indicates that our solution is correct because both sides of the equation are equal.
Checking solutions defends against the possibility of extraneous solutions, which are results that emerge from the algebraic manipulations but don’t hold true in the original equation. It’s a safeguard to ensure that the solutions derived are precise and applicable. In some cases, when you find more than one solution, each must be checked separately to determine which, if any, are true solutions to the original equation.
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