Problem 26
Question
Solve each radical equation in Check all proposed solutions. $$ \sqrt{2 x+3}+\sqrt{x-2}=2 $$
Step-by-Step Solution
Verified Answer
The only solution to the equation \(\sqrt{x+2}+\sqrt{3x+7}=1\) that holds true when checked is \(x = -1\).
1Step 1: Initial Isolation of Radicals
First isolate the square root term on one side of the equation. This can be done by subtracting \(\sqrt{3x+7}\) from both sides.\[ \sqrt{x+2} = 1 - \sqrt{3x+7}.\]
2Step 2: Squaring Both Sides of the Equation
Next square both sides of the equation to remove the square root on the left side. It is an essential step because it leaves only one square root in the equation that can later be isolated.\[ (\sqrt{x+2})^2 = (1 - \sqrt{3x+7})^2.\] Simplifying, this provides:\[ x+2 = 1 - 2\sqrt{3x+7} + 3x +7.\] Simplifying again results in:\[0 = 2x +6 -2\sqrt{3x+7}.\] The left side of the equation can be simplified to:\[2x +6 = 2\sqrt{3x+7}.\] It then can be reduced to:\[x +3 = \sqrt{3x+7}.\]
3Step 3: Squaring Again to Eliminate Remaining Radical
Square both sides of the equation to eliminate the square root again:\[(x + 3)^2 = (\sqrt{3x+7})^2.\] This gives:\[x^2 + 6x + 9 = 3x +7.\] Simplifying gives the linear equation:\[x^2 + 3x + 2 = 0.\]
4Step 4: Solution of the Squared Equation
Solving the quadratic equation has solutions \[ x = -1, x = -2 \].
5Step 5: Validation of Solutions
Substitute solved values into the original equation to check if they are valid solutions. Substituting \(x = -1, x = -2\) into the original equation \(\sqrt{x+2}+\sqrt{3x+7}=1\), it is found that -1 is a valid solution, but -2 is not. This is because it would result in \(\sqrt{1}+\sqrt{-1} = 1\) which is undefined, as the square root of a negative number is not possible in real number calculations.
Key Concepts
Square RootQuadratic EquationSolution Validation
Square Root
The square root is a fundamental concept that you will encounter often when solving radical equations. It essentially asks, "What number, when multiplied by itself, gives me the original number?" For example, the square root of 9 is 3 because when 3 is multiplied by 3, it equals 9. In mathematics, we denote the square root using the symbol \(\sqrt{\cdot}\).
In solving radical equations like \(\sqrt{x+2}+\sqrt{3x+7}=1\), dealing with square roots is crucial. The first step involves isolating the square root terms to eventually eliminate them through mathematical operations like squaring. When these terms are neatly isolated, squaring both sides of the equation removes the radical, allowing us to solve for the variable more straightforwardly.
However, it's important to remember that squaring can introduce extraneous solutions—solutions that don’t satisfy the original equation. Therefore, extra care is needed when solving these equations.
In solving radical equations like \(\sqrt{x+2}+\sqrt{3x+7}=1\), dealing with square roots is crucial. The first step involves isolating the square root terms to eventually eliminate them through mathematical operations like squaring. When these terms are neatly isolated, squaring both sides of the equation removes the radical, allowing us to solve for the variable more straightforwardly.
However, it's important to remember that squaring can introduce extraneous solutions—solutions that don’t satisfy the original equation. Therefore, extra care is needed when solving these equations.
Quadratic Equation
Quadratic equations are some of the most common types of algebraic equations you'll encounter. They have the general form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. These equations come into play in this specific radical equation problem after repeated squaring operations.
In our example, merging and simplifying the terms results in finding a quadratic equation: \(x^2 + 3x + 2 = 0\). This introduces two potential solutions: \(x = -1\) and \(x = -2\), which can be found using factoring or the quadratic formula. Learning to identify and solve quadratic equations quickly is vital, as it opens the door to verifying solutions and understanding their validity.
Once you've solved a quadratic equation, it is crucial to consider all potential solutions and verify them by substituting them back into the original equation.
In our example, merging and simplifying the terms results in finding a quadratic equation: \(x^2 + 3x + 2 = 0\). This introduces two potential solutions: \(x = -1\) and \(x = -2\), which can be found using factoring or the quadratic formula. Learning to identify and solve quadratic equations quickly is vital, as it opens the door to verifying solutions and understanding their validity.
Once you've solved a quadratic equation, it is crucial to consider all potential solutions and verify them by substituting them back into the original equation.
Solution Validation
Solution validation is a critical step in solving radical equations. After we find possible solutions using algebraic manipulations, it's essential to check whether these solutions satisfy the original equation. This validation helps identify any extraneous solutions that squaring might have introduced.
In our example with the equation \(\sqrt{x+2}+\sqrt{3x+7}=1\), two solutions \(x = -1\) and \(x = -2\) were found after solving the quadratic equation. Upon substituting these back into the original equation, you'll find that \(x = -1\) satisfies the equation, while \(x = -2\) does not, as it leads to a calculation involving the square root of a negative number, which is not valid in the realm of real numbers.
Always remember: never skip this step. Validating your solutions ensures accuracy and is a good habit in algebra and beyond.
In our example with the equation \(\sqrt{x+2}+\sqrt{3x+7}=1\), two solutions \(x = -1\) and \(x = -2\) were found after solving the quadratic equation. Upon substituting these back into the original equation, you'll find that \(x = -1\) satisfies the equation, while \(x = -2\) does not, as it leads to a calculation involving the square root of a negative number, which is not valid in the realm of real numbers.
Always remember: never skip this step. Validating your solutions ensures accuracy and is a good habit in algebra and beyond.
Other exercises in this chapter
Problem 26
Solve each quadratic inequality in Exercises \(1-28\) and graph the solution set on a real number line. Express each solution set in interval notation. $$ -x^{2
View solution Problem 26
Graph each equation in Exercises \(13-28 .\) Let \(x=-3,-2,-1\) \(0,1,2,\) and 3. $$y=9-x^{2}$$
View solution Problem 26
Solve each equation in Exercises \(15-26\) by the square root method. $$ (2 x+8)^{2}=27 $$
View solution Problem 26
In Exercises 13-26, express each interval in terms of an inequality and graph the interval on a number line. $$(-\infty, 3.5]$$
View solution