Problem 30
Question
Check all proposed solutions. $$ \sqrt{1+4 \sqrt{x}}=1+\sqrt{x} $$
Step-by-Step Solution
Verified Answer
The solution is that the equation has no valid solutions.
1Step 1: Isolate the square root
Rearrange the equation to isolate the square root on one side of the equation. The equation thus becomes \( \sqrt{1+4 \sqrt{x}} -1 = \sqrt{x}\).
2Step 2: Square both sides
Squaring gets rid of square roots. Square both sides of the equation. However, it's important to remember that squaring only gives the positive square root. This yields \( \left(\sqrt{1+4 \sqrt{x}} -1\right)^2 = (\sqrt{x})^2\) which simplifies to \(1+4\sqrt{x} - 2\sqrt{1+4\sqrt{x}} = x\).
3Step 3: Isolate the square root
Now isolate another square root: \(4\sqrt{x} - 2\sqrt{1+4\sqrt{x}} = x-1\).
4Step 4: Again, square both sides
Squaring gives: \( \left(4\sqrt{x} - 2\sqrt{1+4\sqrt{x}}\right)^2 = (x-1)^2 \), which simplifies to \(16x -16\sqrt{x}+16\sqrt{x}-4\sqrt{(1+4\sqrt{x})^2} = x^2-2x+1. \) This simplifies further to \(16x -4+4\sqrt{x}-\sqrt{x}^2 = x^2-2x+1\).
5Step 5: Simplify and Solve
Finally solving yields the equation x-3 = 0 and so the solution is x=3.
6Step 6: Check the solution
A potential solution according to our manipulation of the equation is x=3. Substitute it back into the initial equation to verify it holds true: \(\sqrt{1+4\sqrt{3}} = 1+\sqrt{3}\). Simplifying the left side we find \(\sqrt{1+4\sqrt{3}} = \sqrt{1+2\sqrt{9}} = \sqrt{1+6}= \sqrt{7}\). The right side is \(1 + \sqrt{3} = \sqrt{4} = 2\). Since \( \sqrt{7} ≠ 2 \), the initial solution is not valid and thus the equation has no solution.
Key Concepts
Squaring Both SidesIsolating the RadicalChecking SolutionsInvalid Solutions
Squaring Both Sides
When solving radical equations, one common method is to square both sides of the equation. This technique helps to eliminate the square root, making it easier to solve the equation. Imagine you have an equation like \(\sqrt{1 + 4 \sqrt{x}} = 1 + \sqrt{x}\). To remove the square root on the left, you square each side. This transforms it to \((\sqrt{1 + 4 \sqrt{x}})^2 = (1 + \sqrt{x})^2\).
This yields \(1+4\sqrt{x} - 2\sqrt{1+4\sqrt{x}} = x\).
Keep in mind that when you square both sides, any negative numbers or complex roots might be ignored, leading to equations that look valid but aren't actually solutions to the original equation.
This yields \(1+4\sqrt{x} - 2\sqrt{1+4\sqrt{x}} = x\).
Keep in mind that when you square both sides, any negative numbers or complex roots might be ignored, leading to equations that look valid but aren't actually solutions to the original equation.
Isolating the Radical
Before you can square both sides of a radical equation, it is a good idea to isolate the radical expression. For example, start with \(\sqrt{1+4 \sqrt{x}} = 1+\sqrt{x}\) and aim to manipulate this equation to get the square root on one side.
Rearranging the equation will help you with this. The equation can be adjusted to become \(\sqrt{1+4 \sqrt{x}} -1 = \sqrt{x}\), where the radical is isolated as much as possible.
This makes it simpler when you square the side with the radical, allowing clearer paths to solving the variable \(x\). It's important to follow this step carefully to simplify subsequent calculations.
Rearranging the equation will help you with this. The equation can be adjusted to become \(\sqrt{1+4 \sqrt{x}} -1 = \sqrt{x}\), where the radical is isolated as much as possible.
This makes it simpler when you square the side with the radical, allowing clearer paths to solving the variable \(x\). It's important to follow this step carefully to simplify subsequent calculations.
Checking Solutions
After finding potential solutions, you must substitute back into the original equation to check whether they really work. This step is critical because the process of squaring both sides can sometimes introduce extraneous solutions. When a particular solution is found, like \(x=3\), don't assume it's correct immediately.
Substitute \(x=3\) back into the original equation \(\sqrt{1+4\sqrt{x}}=1+\sqrt{x}\).
On the left, plug in \(x=3\) and simplify: \(\sqrt{1+4\sqrt{3}}=\sqrt{7}\).
On the right side, it becomes \(1 + \sqrt{3} = 2\). Since \(\sqrt{7} eq 2\), the solution does not hold. Hence, this step is vital to ensure the solutions are valid.
Substitute \(x=3\) back into the original equation \(\sqrt{1+4\sqrt{x}}=1+\sqrt{x}\).
On the left, plug in \(x=3\) and simplify: \(\sqrt{1+4\sqrt{3}}=\sqrt{7}\).
On the right side, it becomes \(1 + \sqrt{3} = 2\). Since \(\sqrt{7} eq 2\), the solution does not hold. Hence, this step is vital to ensure the solutions are valid.
Invalid Solutions
Even after performing all steps correctly, sometimes solutions appear that don't truly satisfy the original equation. These are called invalid or extraneous solutions.
They often arise when certain operations, like squaring both sides, change the nature of the solutions. In our exercise, squaring altered \(\sqrt{1+4\sqrt{x}} = 1+\sqrt{x}\) into a simpler equation, but the solution \(x=3\) didn't work when substituted back.
This highlights the importance of double-checking solutions. These checking steps help filter out any invalid solutions, ensuring only accurate and reasonable answers are considered.
They often arise when certain operations, like squaring both sides, change the nature of the solutions. In our exercise, squaring altered \(\sqrt{1+4\sqrt{x}} = 1+\sqrt{x}\) into a simpler equation, but the solution \(x=3\) didn't work when substituted back.
This highlights the importance of double-checking solutions. These checking steps help filter out any invalid solutions, ensuring only accurate and reasonable answers are considered.
Other exercises in this chapter
Problem 30
Solve each equation in Exercises \(15-34\) by the square root property. $$ (4 x-1)^{2}=16 $$
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Use interval notation to express solution sets and graph each solution set on a number line. Solve each linear inequality. $$8 x-2 \geq 14$$
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Contain linear equations with constants in denominators. Solve equation. \(\frac{3 x}{5}-\frac{x-3}{2}=\frac{x+2}{3}\)
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In Exercises \(29-44\), perform the indicated operations and write the result in standard form. $$ 5 \sqrt{-16}+3 \sqrt{-81} $$
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