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Chapter 7: Problem 29

Solve. Check for extraneous solutions. \((2 x)^{\frac{1}{2}}=(x+5)^{\frac{1}{2}}\)

Short Answer

Expert verified
The solution to the equation \((2 x)^{\frac{1}{2}}=(x+5)^{\frac{1}{2}}\) is \(x=5\).

Step by step solution

01

Square both sides of the equation

Start by squaring both sides of the equation to get rid of the square root: \((2x)^{1/2^2}=(x+5)^{1/2^2}\). This simplifies to \(2x = x+5\).
02

Simplify and solve for \(x\)

In order to gather all terms involving \(x\) on one side and constants on the other, subtract \(x\) from both sides of the equation: \(2x-x = x-x+5\). This results in \(x=5\).
03

Check for extraneous solutions

Substitute the solution \(x=5\) back into the original equation: \((2*(5))^{\frac{1}{2}}= ((5) +5)^{\frac{1}{2}}\). This simplifies to \( (10)^{\frac{1}{2}} = (10)^{\frac{1}{2}}\), which means \(x=5\) is a valid solution to the equation, not an extraneous solution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Extraneous Solutions
When solving radical equations, it's crucial to check solutions for extraneous roots. Extraneous solutions are those that emerge during the solving process but do not actually satisfy the original equation. They often appear when both sides of an equation are squared鈥攁 common step to eliminate radicals.

What are extraneous solutions, and why do they occur?
  • These solutions are not valid for the initial conditions of the problem.
  • They typically arise from algebraic manipulations, like squaring both sides, which can introduce solutions not present initially.
To avoid concluding with incorrect answers, always substitute potential solutions back into the original equation to verify their validity. This ensures any extraneous solutions are identified and discarded.
Squaring Both Sides
A common technique to solve radical equations involves squaring both sides to eliminate the radicals. For the equation \((2x)^{1/2} = (x+5)^{1/2}\), this method helps in simplifying the equation.
  • When each side of the equation is squared, the radicals on both sides disappear.
  • This transforms the equation into a linear form, making it easier to handle using basic algebra.
However, remember that while squaring can simplify, it may also introduce non-existent solutions, hence the need for subsequent verification.
Checking Solutions
Once a potential solution is obtained from the simplified equation, it must be checked in the original equation to ensure its accuracy. For instance, after solving \(2x = x + 5\) to find \(x = 5\), the solution is substituted back into the original equation.
  • If the original equation holds true, then the solution is valid.
  • This step helps in identifying and discarding any extraneous solutions.
By substituting \(x=5\) back, both sides equal \((10)^{1/2}\), confirming \(x = 5\) is accurate and not extraneous.
Simplifying Equations
Simplifying equations is a fundamental step after eliminating radicals. It involves reordering and gathering all like terms. In our example, we move from \(2x = x + 5\) to isolate \(x\) by subtracting \(x\) from both sides.
  • This process makes complex equations more accessible and easier to solve.
  • By simplifying to \(x = 5\), we usually find clear and concrete solutions.
Simplification not only leads to solutions but also helps in checking and verifying their validity, ensuring each step aligns with the original equation's requirements.

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