Complete the following statement.
The radical equation $2+\sqrt{2 x-3}=\sqrt{x+7}$ has a solution set ( $x=$ $\square$ ) and an extraneous root $x=$ $\square$
Answer (2)
Isolate a square root term on one side of the equation.
Square both sides to eliminate the square root and simplify.
Repeat the isolation and squaring process if necessary.
Solve the resulting equation (usually a polynomial equation).
Check all solutions in the original equation to identify extraneous roots. The solution set is { 2 } and the extraneous root is 42 .
Explanation
Problem Analysis We are given the radical equation 2 + 2 x − 3 = x + 7 . Our goal is to find the solution set and any extraneous roots.
Isolating the Square Root First, we isolate one of the square roots. In this case, let's isolate x + 7 on one side: x + 7 = 2 + 2 x − 3
Squaring Both Sides Next, we square both sides of the equation to eliminate the square root: ( x + 7 ) 2 = ( 2 + 2 x − 3 ) 2 Expanding both sides, we get: x + 7 = 4 + 4 2 x − 3 + 2 x − 3 Simplifying, we have: x + 7 = 2 x + 1 + 4 2 x − 3
Isolating the Remaining Square Root Now, we isolate the remaining square root term: 6 − x = 4 2 x − 3
Squaring Both Sides Again We square both sides again to eliminate the square root: ( 6 − x ) 2 = ( 4 2 x − 3 ) 2 Expanding and simplifying: 36 − 12 x + x 2 = 16 ( 2 x − 3 ) 36 − 12 x + x 2 = 32 x − 48
Rearranging into a Quadratic Equation Rearranging into a quadratic equation: x 2 − 44 x + 84 = 0
Solving the Quadratic Equation We solve the quadratic equation for x by factoring: ( x − 2 ) ( x − 42 ) = 0 So, x = 2 or x = 42 .
Checking for Valid and Extraneous Solutions Now, we check both solutions in the original equation to determine if they are valid or extraneous.
For x = 2 : 2 + 2 ( 2 ) − 3 = 2 + 1 = 2 + 1 = 3 2 + 7 = 9 = 3 Since 3 = 3 , x = 2 is a valid solution.
For x = 42 : 2 + 2 ( 42 ) − 3 = 2 + 81 = 2 + 9 = 11 42 + 7 = 49 = 7 Since 11 = 7 , x = 42 is an extraneous root.
Final Answer Therefore, the solution set is { 2 } and the extraneous root is 42 .
Examples
Radical equations are useful in various fields such as physics and engineering. For example, in physics, the period of a pendulum can be modeled using a radical equation. Suppose we have a pendulum whose period T is given by T = 2 π g L , where L is the length of the pendulum and g is the acceleration due to gravity. If we want to find the length L required for a specific period T , we would need to solve this radical equation for L . Similarly, engineers use radical equations to model various physical phenomena, such as fluid flow and electrical circuits. Understanding how to solve these equations is crucial for making accurate predictions and designing effective systems.
The radical equation has a solution set of { 2 } and an extraneous root of 42 .
;
