Problem 25
Question
Solve. Check for extraneous solutions. \(\sqrt{x+7}-x=1\)
Step-by-Step Solution
Verified Answer
The solution to the equation \(\sqrt{x+7}-x=1\) is \(x = 2\).
1Step 1: Isolate the Square Root
First, add \(x\) to both sides of the equation \(\sqrt{x+7}-x=1\) to get \(\sqrt{x+7} = x+1\). This form of the equation allows to square both sides in the next step.
2Step 2: Square Both Sides of the Equation
After step 1, square both sides of the equation to eliminate the square root: squaring \(\sqrt{x+7}\) results in \(x+7\), and squaring \(x+1\) results in \(x^2+2x+1\). Therefore, the equation turns into \(x+7 = x^2+2x+1\).
3Step 3: Simplify the Equation
Group all terms on one side by subtracting \(x\) and \(7\) from both sides: the equation simplifies to \(x^2 + x - 6 = 0\). This result is a simple quadratic equation.
4Step 4: Factor the Quadratic
Factor the quadratic \(x^2 + x - 6\) as \((x - 2)(x + 3) = 0\) to find the solutions.
5Step 5: Solve for x
Set each factor equal to zero and solve for \(x\). We obtain \(x = 2\) and \(x = -3\) as possible solutions.
6Step 6: Check for Extraneous Solutions
Plug the possible solutions back into the original equation to verify if they hold true. The solutions \(x = 2\) holds true since \(\sqrt{2+7} - 2 = 1, whereas x = -3 does not hold true since \(\sqrt{-3+7} - (-3)\neq1. Therefore, the latter is an extraneous solution.
Key Concepts
Quadratic EquationsFactoring QuadraticsSquare Root EquationsChecking Solutions
Quadratic Equations
Quadratic equations are fundamental concepts in algebra. They are equations of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(a eq 0\). These equations represent a parabola when graphed, and they can have zero, one, or two real solutions.
Quadratic equations can be solved using different methods, including:
Quadratic equations can be solved using different methods, including:
- Factoring
- Completing the square
- Using the quadratic formula
Factoring Quadratics
Factoring is one of the simplest methods for solving quadratic equations. It involves writing the quadratic expression as a product of two binomials. This is done by reversing the process of multiplying two binomials (like \((x+p)(x+q)\)).
For example, consider the quadratic equation \(x^2+x-6=0\). We need to find two numbers that multiply to \(-6\) (the constant term) and add to \(1\) (the coefficient of \(x\)). The numbers \(-2\) and \(+3\) fit these criteria, so the equation can be factored into \((x-2)(x+3)=0\).
For example, consider the quadratic equation \(x^2+x-6=0\). We need to find two numbers that multiply to \(-6\) (the constant term) and add to \(1\) (the coefficient of \(x\)). The numbers \(-2\) and \(+3\) fit these criteria, so the equation can be factored into \((x-2)(x+3)=0\).
- Setting each factor equal to zero gives possible solutions.
- Solving \(x-2=0\) yields \(x=2\), and solving \(x+3=0\) yields \(x=-3\).
Square Root Equations
Square root equations contain the variable inside a square root. Solving them often requires isolating the square root on one side before eliminating it by squaring both sides. This process can introduce extraneous solutions, which are solutions that do not satisfy the original equation.
In the exercise, we started with the equation \(\sqrt{x+7}-x=1\). By isolating the square root to get \(\sqrt{x+7}=x+1\), we could then square both sides to remove the square root. Squaring transforms the equation by raising each side to the power of two, making it easier to solve.
In the exercise, we started with the equation \(\sqrt{x+7}-x=1\). By isolating the square root to get \(\sqrt{x+7}=x+1\), we could then square both sides to remove the square root. Squaring transforms the equation by raising each side to the power of two, making it easier to solve.
- Always remember to check your final solutions against the original equation to ensure they are valid.
Checking Solutions
Even after finding solutions through methods like factoring, it's crucial to check each solution in the original equation, especially for equations that involve square roots or other operations that could introduce extraneous solutions.
When solving \(\sqrt{x+7}-x=1\), substituting the potential solutions \(x=2\) and \(x=-3\) reveals that only \(x=2\) is valid. Plugging \(x=2\) into the original equation results in equality \(\sqrt{2+7}-2=1\), confirming it as a genuine solution.
When solving \(\sqrt{x+7}-x=1\), substituting the potential solutions \(x=2\) and \(x=-3\) reveals that only \(x=2\) is valid. Plugging \(x=2\) into the original equation results in equality \(\sqrt{2+7}-2=1\), confirming it as a genuine solution.
- Always test each solution:\
- For \(x=-3\), the equation \(\sqrt{-3+7}-(-3)\) does not yield 1, showing it as extraneous.
Other exercises in this chapter
Problem 25
Solve each square root equation by graphing. Round the answer to the nearest hundredth if necessary. If there is no solution, explain why. \(\sqrt{x-3}=12\)
View solution Problem 25
For each function \(f,\) find \(f^{-1}\) and the domain and range of \(f\) and \(f^{-1} .\) Determine whether \(f^{-1}\) is a function. $$ f(x)=\sqrt{x+7} $$
View solution Problem 25
Let \(g(x)=2 x\) and \(h(x)=x^{2}+4 .\) Evaluate each expression. $$ (g \circ h)(-2) $$
View solution Problem 25
Rationalize each denominator. Simplify the answer. $$ \frac{5+\sqrt{3}}{2-\sqrt{3}} $$
View solution