Problem 11
Question
Solve the equation. Check for extraneous solutions. $$-5+\sqrt{x}=0$$
Step-by-Step Solution
Verified Answer
The solution to the equation is \(x = 25\).
1Step 1: Isolate the square root term
Rearrange the equation to get the square root on one side by itself. To achieve this, add 5 to both sides of the equation. The resulting equation will be \(\sqrt{x}=5\)
2Step 2: Eliminate the square root
To get rid of the square root, square both sides of the equation obtaining \(x = 25\)
3Step 3: Check for extraneous solutions
Substitute \(x = 25\) back into the original equation to see if it satisfies the equation. After replacement, the left side of the equation becomes \(-5 + \sqrt{25}\), which simplifies to \(0\), which is equal to the right side. Therefore, x = 25 is not an extraneous solution.
Key Concepts
Extraneous SolutionsIsolating the Square RootSquaring Both Sides
Extraneous Solutions
In the realm of algebra, an extraneous solution is a number that emerges as a possible solution to an equation during the solving process but does not actually satisfy the original equation upon verification.
Extraneous solutions often appear when both sides of an equation are raised to a power, as this action can introduce results that weren't solutions for the initial equation. For example, when solving square root equations, after squaring both sides, it's crucial to plug the solution back into the original equation to confirm its validity.
Checking for extraneous solutions is like doing quality control, ensuring that the answer found is truly a part of the solution set. This extra step is important because it prevents incorrect solutions from being accepted, thus keeping the integrity of our mathematical conclusions intact.
Extraneous solutions often appear when both sides of an equation are raised to a power, as this action can introduce results that weren't solutions for the initial equation. For example, when solving square root equations, after squaring both sides, it's crucial to plug the solution back into the original equation to confirm its validity.
Checking for extraneous solutions is like doing quality control, ensuring that the answer found is truly a part of the solution set. This extra step is important because it prevents incorrect solutions from being accepted, thus keeping the integrity of our mathematical conclusions intact.
Isolating the Square Root
The technique of isolating the square root on one side of the equation is a key step in solving square root equations. It sets the stage for eliminating the square root altogether.
To isolate the square root, we move all other terms to the opposite side of the equation. This can be done through basic algebraic operations like addition or subtraction. In our example, we added 5 to each side to get \(\sqrt{x}=5\).
To isolate the square root, we move all other terms to the opposite side of the equation. This can be done through basic algebraic operations like addition or subtraction. In our example, we added 5 to each side to get \(\sqrt{x}=5\).
Why Isolate the Square Root?
By isolating the square root, we're able to focus on the core element that requires manipulation, making the remaining steps of the solving process more straightforward. Moreover, it allows us to apply the inverse operations effectively in the next step.Squaring Both Sides
Once the square root is isolated, our subsequent action is squaring both sides. This maneuver is the key to unlocking the variable within the square root.
By squaring both sides, we apply the principle that the square of a square root returns the original value under the root, thus eliminating the radical itself—\(\sqrt{x}^2 = x\). It's an elegant undoing of the square root.
By squaring both sides, we apply the principle that the square of a square root returns the original value under the root, thus eliminating the radical itself—\(\sqrt{x}^2 = x\). It's an elegant undoing of the square root.
Why Squaring Matters
The importance of this step lies in its power to transform. With it, we can change a square root equation into a polynomial equation, usually linear or quadratic, for which we have well-established solving techniques. Once the equation is squared and the root eliminated, it's simpler to arrive at the solution, as shown in our example, where we easily find that \(x=25\).Other exercises in this chapter
Problem 10
Solve by completing the square. $$x^{2}-2 x-18=0$$
View solution Problem 10
Simplify the expression. $$\frac{4}{\sqrt{13}}$$
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State the basic axiom of algebra that is represented. $$(x+y)+z=x+(y+z)$$
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Find the midpoint between the two points. \((6,2),(2,-3)\)
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