Problem 24
Question
Solve each radical equation. Don't forget, you must check potential solutions. $$\sqrt{x}=x-2$$
Step-by-Step Solution
Verified Answer
The solution is \(x = 4\).
1Step 1: Square Both Sides
To eliminate the square root, square both sides of the equation: \[(\sqrt{x})^2 = (x - 2)^2\] which simplifies to \[x = x^2 - 4x + 4\].
2Step 2: Rearrange the Equation
Move all terms to one side to set the equation to zero: \[x^2 - 4x + 4 - x = 0\], which simplifies to \[x^2 - 5x + 4 = 0\].
3Step 3: Factor the Quadratic Equation
Factor the quadratic equation: \[x^2 - 5x + 4 = (x - 4)(x - 1) = 0\].
4Step 4: Solve for x
Set each factor to zero and solve for \(x\):\[x - 4 = 0 \quad \Rightarrow \quad x = 4\] \[x - 1 = 0 \quad \Rightarrow \quad x = 1\].
5Step 5: Check Potential Solutions
Substitute \(x = 4\) back into the original equation: \(\sqrt{4} = 4 - 2\), which becomes \(2 = 2\). Substitute \(x = 1\) back into the original equation: \(\sqrt{1} = 1 - 2\), which becomes \(1 = -1\), which is false. Thus, \(x = 1\) is not a solution.
Key Concepts
Checking SolutionsSquaring Both SidesFactoring QuadraticsPotential Extraneous Solutions
Checking Solutions
After solving a radical equation, it is crucial to verify the solutions by substituting them back into the original equation. This step ensures the solutions actually satisfy the equation. When we substitute a solution and it holds true, it confirms the validity of the solution. But if it doesn't, it means the solution is not correct and is known as an extraneous solution.
This happens because of the nature of dealing with square roots and squares. In our example, substituting \(x = 4\) back into \(\sqrt{x} = x - 2\) gives us \(2 = 2\), which is true, confirming that \(x = 4\) is a solution.
On the other hand, substituting \(x = 1\) results in \(1 = -1\), which does not hold true. Thus, \(x = 1\) is not a valid solution.
This happens because of the nature of dealing with square roots and squares. In our example, substituting \(x = 4\) back into \(\sqrt{x} = x - 2\) gives us \(2 = 2\), which is true, confirming that \(x = 4\) is a solution.
On the other hand, substituting \(x = 1\) results in \(1 = -1\), which does not hold true. Thus, \(x = 1\) is not a valid solution.
Squaring Both Sides
Solving radical equations often involves eliminating the square root sign to simplify the equation further. We do this by squaring both sides of the equation. Squaring both sides ensures that all terms are easily manageable, especially when it comes to further algebraic manipulation.
In our case, with \(\sqrt{x} = x - 2\), squaring provides us with \(x = (x - 2)^2\). It's important to perform this operation carefully, as it must be done correctly to obtain the right form of the equation for further simplification.
Always remember that squaring can sometimes introduce extraneous solutions, which is why checking solutions later is so crucial.
In our case, with \(\sqrt{x} = x - 2\), squaring provides us with \(x = (x - 2)^2\). It's important to perform this operation carefully, as it must be done correctly to obtain the right form of the equation for further simplification.
Always remember that squaring can sometimes introduce extraneous solutions, which is why checking solutions later is so crucial.
Factoring Quadratics
Once the square root has been eliminated, you may end up with a quadratic equation. The next step involves rearranging the equation and factoring the quadratic expression.
Factoring is the process of breaking down the quadratic into simpler, solvable terms. Specifically, you rewrite the quadratic into a product of binomials: \(x^2 - 5x + 4 = (x - 4)(x - 1) = 0\).
This is a straightforward approach to finding solutions. By setting each factor equal to zero, you solve for \(x\). Factoring allows you to find potential solutions that will need to be checked against the original equation.
Factoring is the process of breaking down the quadratic into simpler, solvable terms. Specifically, you rewrite the quadratic into a product of binomials: \(x^2 - 5x + 4 = (x - 4)(x - 1) = 0\).
This is a straightforward approach to finding solutions. By setting each factor equal to zero, you solve for \(x\). Factoring allows you to find potential solutions that will need to be checked against the original equation.
Potential Extraneous Solutions
An extraneous solution is a solution derived from the algebraic process that does not satisfy the original equation. These are solutions that appear valid but are not.
Potential extraneous solutions often arise when both sides of a radical equation are squared. That's why checking solutions is so important. In our example, when we solve \((x - 4)(x - 1) = 0\), we get \(x = 4\) and \(x = 1\). However, checking these shows only \(x = 4\) satisfies the original \(\sqrt{x} = x - 2\).
This means \(x = 1\) is an extraneous solution. Remember, it's essential not to assume all derived solutions are correct without checking them in the original equation.
Potential extraneous solutions often arise when both sides of a radical equation are squared. That's why checking solutions is so important. In our example, when we solve \((x - 4)(x - 1) = 0\), we get \(x = 4\) and \(x = 1\). However, checking these shows only \(x = 4\) satisfies the original \(\sqrt{x} = x - 2\).
This means \(x = 1\) is an extraneous solution. Remember, it's essential not to assume all derived solutions are correct without checking them in the original equation.
Other exercises in this chapter
Problem 24
Use the quadratic formula to solve each of the quadratic equations. Check your solutions by using the sum and product relationships. $$2 x^{2}+5 x-2=0$$
View solution Problem 24
Use the method of completing the square to solve each quadratic equation. $$n(n+14)=-4$$
View solution Problem 24
Add or subtract as indicated. $$\left(\frac{2}{3}-\frac{1}{5} i\right)+\left(\frac{3}{5}-\frac{3}{4} i\right)$$
View solution Problem 25
Solve each inequality. $$3 x^{2}+13 x-10 \leq 0$$
View solution