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Chapter 8: Problem 27

Solve each equation. $$ \sqrt{5 x-5}=\sqrt{4 x+1} $$

Short Answer

Expert verified
The solution is x = 6.

Step by step solution

01

Isolate the square roots

Notice that both sides of the equation contain a square root. Since they are equal, you can set the contents of the square roots equal to each other: \[ 5x - 5 = 4x + 1 \]
02

Solve for x

Now, solve for the variable x by first getting all the x-terms on one side of the equation: \[ 5x - 4x = 1 + 5 \] This simplifies to: \[ x = 6 \]
03

Verify the solution

Substitute the value of x back into the original equation to verify it. First, check the left side: \[ \text{Left side} = \sqrt{5(6) - 5} = \sqrt{30 - 5} = \sqrt{25} = 5 \] Then, check the right side: \[ \text{Right side} = \sqrt{4(6) + 1} = \sqrt{24 + 1} = \sqrt{25} = 5 \] Both sides are equal, so the solution is verified.

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Key Concepts

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

Isolating Square Roots
When solving equations that involve square roots, the first important step is to isolate the square roots. This means ensuring that the square root expression stands alone on one side of the equation.
In our example equation \(umerator \sqrt{5 x - 5} = \umerator \sqrt{4 x + 1}\), we already have square roots isolated on both sides. Therefore, we can equate the contents within the square roots directly:
  • Set \(5x - 5\) equal to \(4x + 1\): \[5x - 5 = 4x + 1\]
This step is crucial because it simplifies the equation and removes the square roots, making it easier to solve for the variable.
Solving for a Variable
The next step is solving for the variable, in this case, x. Begin by rearranging the equation so that all terms involving x are on one side and the constants on the other.
From the isolated equation \[5x - 5 = 4x + 1\], subtract 4x from both sides to collect x-terms:
  • \(5x - 4x = 1 + 5\)
This simplifies to:
  • \(x = 6\)
Now, we have isolated x and found its value. Always make sure your equation is as simple as possible before solving for the variable.
Verifying Solutions in Algebra
The final and essential step is to verify the found solution by substituting it back into the original equation. This ensures accuracy and confirms that no mistakes were made.
Let's substitute x = 6 back into the initial equation \(umerator \sqrt{5 x - 5} = \umerator \sqrt{4 x + 1}\). First, check the left side:
  • \(\sqrt{5(6) - 5} = \sqrt{30 - 5} = \sqrt{25} = 5\)
Now check the right side:
  • \(\sqrt{4(6) + 1} = \sqrt{24 + 1} = \sqrt{25} = 5\)
Since both sides are equal (5 = 5), the solution x = 6 is verified to be correct. This step confirms the integrity of your solution and ensures that it fits the original equation.

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