Problem 77
Question
Solve each equation. Write all proposed solutions. Cross out those that are extraneous. $$ \sqrt{y+2}+y=4 $$
Step-by-Step Solution
Verified Answer
The solution is \( y = 2 \); \( y = 7 \) is extraneous.
1Step 1: Isolate the Square Root Term
Start by isolating the square root term on one side of the equation. We have: \( \sqrt{y+2} + y = 4 \). Subtract \( y \) from both sides to get: \( \sqrt{y+2} = 4 - y \).
2Step 2: Square Both Sides to Eliminate the Square Root
Square both sides of the equation to eliminate the square root. This gives us: \( (\sqrt{y+2})^2 = (4 - y)^2 \). Which simplifies to: \( y + 2 = 16 - 8y + y^2 \).
3Step 3: Rearrange into a Standard Quadratic Equation
Bring all terms to one side to form a quadratic equation: \( y^2 - 9y + 14 = 0 \).
4Step 4: Solve the Quadratic Equation
Factor the quadratic equation: \( (y - 7)(y - 2) = 0 \). This yields the potential solutions: \( y = 7 \) and \( y = 2 \).
5Step 5: Verify the Solutions for Extraneous Results
Check each solution in the original equation to verify whether it is an actual solution to the equation. - Substitute \( y = 7 \) into the original equation: \( \sqrt{7 + 2} + 7 = \sqrt{9} + 7 = 3 + 7 = 10 eq 4 \). So, \( y = 7 \) is extraneous. - Substitute \( y = 2 \) into the original equation: \( \sqrt{2 + 2} + 2 = \sqrt{4} + 2 = 2 + 2 = 4 \). So, \( y = 2 \) is a valid solution.
Key Concepts
Square RootExtraneous SolutionsSolving Equations
Square Root
The square root function can be quite the head-scratcher when seen in equations. Let's break it down. The square root of a number refers to a value that, when multiplied by itself, gives the original number. For example, the square root of 9 is 3, because 3 times 3 equals 9. In math notation, we write this as \( \sqrt{9} = 3 \).
In equations, the square root often needs our full attention. We must isolate it first before moving on to other operations. In our problem, we began with \( \sqrt{y+2} + y = 4 \). By isolating \( \sqrt{y+2} \), we simplified the next steps. This approach makes it easier to work with and ultimately tackle the entire problem correctly.
In equations, the square root often needs our full attention. We must isolate it first before moving on to other operations. In our problem, we began with \( \sqrt{y+2} + y = 4 \). By isolating \( \sqrt{y+2} \), we simplified the next steps. This approach makes it easier to work with and ultimately tackle the entire problem correctly.
- Isolation reduces complexity.
- Always check your square roots, as both positive and negative roots can sometimes present valid solutions.
Extraneous Solutions
Extraneous solutions are tricky! They are solutions that appear when solving an equation but aren’t really solutions to the original equation. This often happens when we perform operations like squaring both sides of an equation. Why does this occur? Because squaring can introduce solutions that don't truly satisfy the start of our problem.
In our problem, after rearranging and solving, we got two possible solutions: \( y = 7 \) and \( y = 2 \). However, upon checking back with the original equation \( \sqrt{y+2} + y = 4 \), the solution \( y = 7 \) turned out to produce a false statement, making it extraneous.
In our problem, after rearranging and solving, we got two possible solutions: \( y = 7 \) and \( y = 2 \). However, upon checking back with the original equation \( \sqrt{y+2} + y = 4 \), the solution \( y = 7 \) turned out to produce a false statement, making it extraneous.
- Check all proposed solutions by substituting them back into the initial equation.
- Being thorough avoids incorrect conclusions in your final answer.
Solving Equations
Solving equations is all about finding values for variables that make the equation true. The journey involves various steps, depending on the equation's complexity. Let's map this out:
The equation from our problem \( \sqrt{y+2} + y = 4 \) needed simplification before solutions could emerge. After isolating the square root and transforming our equation through squaring both sides, we arranged terms into a neat quadratic form: \( y^2 - 9y + 14 = 0 \).
To solve this quadratic equation, we factored it to get \( (y - 7)(y - 2) = 0 \), leading to potential solutions. Remember, sometimes other methods such as quadratic formula or completing the square might be necessary, yet factoring is often the simplest path if applicable.
The equation from our problem \( \sqrt{y+2} + y = 4 \) needed simplification before solutions could emerge. After isolating the square root and transforming our equation through squaring both sides, we arranged terms into a neat quadratic form: \( y^2 - 9y + 14 = 0 \).
To solve this quadratic equation, we factored it to get \( (y - 7)(y - 2) = 0 \), leading to potential solutions. Remember, sometimes other methods such as quadratic formula or completing the square might be necessary, yet factoring is often the simplest path if applicable.
- Always ensure the equation is simplified as much as possible first.
- Apply appropriate algebraic techniques methodically.
- Double-check solutions to ensure no extraneous surprises.
Other exercises in this chapter
Problem 76
Simplify each expression, if possible. All variables represent positive real numbers. $$ 3 \sqrt[4]{x^{4} y}-2 \sqrt[4]{x^{4} y} $$
View solution Problem 77
Simplify each cube root. See Example 6. $$ \sqrt[3]{64} $$
View solution Problem 77
Simplify each expression. Write the answers without negative exponents. All variables represent positive real numbers. See Example 8. $$ 9^{3 / 7} \cdot 9^{2 /
View solution Problem 77
Divide. Write all answers in the form a \(+b i.\) $$ \frac{7+3 i}{4-2 i} $$
View solution