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Chapter 7: Problem 28

Solve. Check for extraneous solutions. \(x+8=\left(x^{2}+16\right)^{\frac{1}{2}}\)

Short Answer

Expert verified
The solution to the equation is \(x = -3\).

Step by step solution

01

Squaring both sides

The square root on the right side limits the operations that can be performed on this equation. That is why squaring both sides is the next logical step. After squaring, our equation will look like this: \( (x+8)^2 = x^2 + 16\). The equation expands to \(x^2 + 16x + 64 = x^2 + 16\).
02

Simplifying the equation

Now, simplify the equation. To do that, subtract \(x^2\) and 16 from both sides: \(16x + 64 = 16\). Further simplify it to \(16x = -48\) and then \( x = -48 / 16\), which gives you \(x = -3\).
03

Checking for extraneous solutions

Extraneous solutions can often occur due to the process of squaring both sides, so it's important to check. Substitute \(x = -3\) back into the original equation, \(x+8=\left(x^{2}+16\right)^{\frac{1}{2}}\), to get \(-3+8=\left((-3)^{2}+16\right)^{\frac{1}{2}}\). Simplify both sides to get 5 = 5, which proves that \(x = -3\) is a valid solution and not an extraneous solution.

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Key Concepts

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

Solving Equations
When solving equations, the aim is to find the value of the variable that makes the equation true.
Solving equations often involves transformations to simplify or rearrange the given expression.
These steps might include basic operations such as addition, subtraction, multiplication, or division.In this exercise, the equation given is \(x+8=\left(x^{2}+16\right)^{\frac{1}{2}}\).
The goal is to isolate \(x\) and solve for its value.
This process requires careful manipulation of the equation to preserve equality and ensure accuracy.
  • Transformations must be applied equally to both sides of the equation to maintain balance.
  • The presence of the square root in this equation makes direct isolation of \(x\) a bit more complex.
Choosing the right method to tackle an equation like this is crucial and often involves intuition and experience.
Squaring Both Sides
Squaring both sides is a useful technique when the equation involves a square root, as it allows us to eliminate the square root and work with a polynomial equation.
By squaring the equation \(x+8=\left(x^{2}+16\right)^{\frac{1}{2}}\), we remove the square root symbol, transforming it into \( (x+8)^2 = x^2 + 16\).Let's break down the process:
  • First, ensure both sides of the equation are squared correctly. In our case, the left side becomes \(x^2 + 16x + 64\) after expanding \((x+8)^2\).
  • This step facilitates getting rid of the square root, which simplifies further mathematics.
Squaring, while useful, can also introduce extraneous solutions, which means checking solutions afterwards is important.
Checking Solutions
Checking solutions is a critical step to verify the validity of the solutions found, especially after operations like squaring both sides.
An extraneous solution is one that arises from the process of solving the equation but doesn't satisfy the original equation.In the given example, once we found \(x = -3\), we substitute it back into the original equation: \((-3)+8=\left((-3)^{2}+16\right)^{\frac{1}{2}}\), simplifying to \(5 = 5\).
  • This confirms \(x = -3\) is indeed a solution and not extraneous.
  • When checking solutions, always substitute back into the original equation rather than the transformed one.
This extra verification step assures that the solution is not only mathematically derived but also correct in context.

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Most popular questions from this chapter

Let \(f(x)=4 x, g(x)=\frac{1}{2} x+7,\) and \(h(x)=|-2 x+4| .\) Simplify each function. $$ (h \circ(g \circ f))(x) $$

Find the inverse of each function. Is the inverse a function? \(f(x)=2 x^{3}\)

Graph. Find the domain and the range of each function. \(y=7-\sqrt{2 x-1}\)

Graph each function. \(y=2 \sqrt[3]{x-6}-9\)

Find each indicated root if it is a real number. $$ \sqrt[4]{16} $$
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