Problem 59 - Algebra and Trigonometry

Question

Solve:$$\sqrt{x}+\sqrt{x-5}=5$$

Step-by-Step Solution

Answer

The solution to the equation is \(x = 9\).

Step 1: Isolate One Square Root

To begin, isolate one of the square roots. To do this, subtract \(\sqrt{x-5}\) from both sides of the given equation \(\sqrt{x}+\sqrt{x-5}=5\) to obtain the equation \(\sqrt{x}=5-\sqrt{x-5}\).

Step 2: Square Both Sides of The Equation

To get rid of the square roots, we can square each side of the equation. This gives, \[ (\sqrt{x})^2 = (5 - \sqrt{x-5})^2 \]. Simplifying, this leads to \[ x = 25 - 10\sqrt{x-5} + x - 5 \]. Simplifying this equation further, we get \[ 0 = 20 - 10\sqrt{x-5} \].

Step 3: Isolate the Remaining Square root

To further isolate the remaining square root, it's convenient to firstly add 20 to both sides to get \[ 10\sqrt{x-5} = 20 \].

Step 4: Solve for x

Now, divide both sides of the equation \[ 10\sqrt{x-5} = 20 \] by 10. Hence, the equation simplifies to \[ \sqrt{x-5} = 2 \]. Again, square both sides to get \[ x - 5 = 4 \], which results in the value \[ x = 9 \].

Step 5: Checking the solution

It is always important to verify the solution found. Substituting the value \( x = 9 \) into the original equation shows that this value of \(x\) is indeed a valid solution.\[ \sqrt{9}+\sqrt{9-5} = 5 \]\[ 3 + 2 = 5 \] is a valid equation, so \(x = 9\) is indeed a valid solution.

Key Concepts

Isolating Square Roots

When you have an equation with square roots, like \( \sqrt{x} + \sqrt{x-5} = 5 \), it's important to start by isolating one of these square roots. This simplification helps manage the components of the equation individually. Start by moving one square root to the other side of the equation. Here, we subtract \( \sqrt{x-5} \) from both sides.

• This gives us \( \sqrt{x} = 5 - \sqrt{x-5} \).

Once this is achieved, you've simplified the equation into a manageable form where one square root is isolated. This makes it easier to perform further algebraic manipulations.

Verify Solution

Verifying solutions is a crucial step in solving equations, especially when square roots are involved. Once you've found a value for \( x \), you must check that it satisfies the original equation. Mistakes can happen during the solving process, particularly when dealing with squaring operations and square roots.
For our solution, we substitute \( x = 9 \) back into the original equation:

• Calculate \( \sqrt{9} = 3 \) and \( \sqrt{9-5} = \sqrt{4} = 2 \).

This gives us \( 3 + 2 = 5 \), which is true. Hence, \( x = 9 \) is verified, ensuring the solution is accurate and reliable.

Algebraic Manipulation

Algebraic manipulation refers to the various mathematical techniques used to rewrite and solve equations. In this exercise, a key step involves reformulating the equation after isolating a square root. Once simplified, you start manipulating terms so you can eliminate the square roots.

• Like adding or subtracting terms from both sides of the equation to simplify or isolate variables.
• Example: After isolating \( \sqrt{x} \), you work towards eliminating \( \sqrt{x-5} \).

Algebraic manipulation helps shape equations into forms that are easier to solve, leading you step by step towards the solution.

Square Both Sides

Squaring both sides of an equation is an essential technique when working with square roots. Once you isolate a square root, the next logical step is to square both sides to eliminate it:

• For instance, squaring \( \sqrt{x} = 5 - \sqrt{x-5} \) results in \( x = (5 - \sqrt{x-5})^2 \).
• This changes the equation's form and simplifies the problem, removing the square root so standard algebraic techniques can be applied.

Caution is needed, as this operation can introduce extraneous solutions. Therefore, checking your solutions by substituting back into the original equation remains essential.