Solve the equation: [tex]$\sqrt{x+11}=x+5$[/tex]
Answer (1)
Square both sides of the equation x + 11 = x + 5 to get x + 11 = ( x + 5 ) 2 .
Expand and rearrange to form the quadratic equation x 2 + 9 x + 14 = 0 .
Factor the quadratic equation to get ( x + 2 ) ( x + 7 ) = 0 , which gives potential solutions x = − 2 and x = − 7 .
Check the solutions in the original equation to find that x = − 2 is a valid solution, while x = − 7 is extraneous. Thus, the final answer is − 2 .
Explanation
Understanding the Problem We are given the equation x + 11 = x + 5 . Our goal is to find the value(s) of x that satisfy this equation.
Squaring Both Sides To eliminate the square root, we square both sides of the equation: ( x + 11 ) 2 = ( x + 5 ) 2 This simplifies to: x + 11 = x 2 + 10 x + 25
Rearranging into Quadratic Form Now, we rearrange the equation to form a quadratic equation: x 2 + 10 x + 25 − x − 11 = 0 x 2 + 9 x + 14 = 0
Factoring the Quadratic We can factor the quadratic equation as follows: ( x + 2 ) ( x + 7 ) = 0
Finding Potential Solutions This gives us two possible solutions for x : x + 2 = 0 ⇒ x = − 2 x + 7 = 0 ⇒ x = − 7
Checking for Extraneous Solutions Now, we need to check if these solutions are valid by substituting them back into the original equation. Let's start with x = − 2 : − 2 + 11 = − 2 + 5 9 = 3 3 = 3 Since this is true, x = − 2 is a valid solution. Now let's check x = − 7 : − 7 + 11 = − 7 + 5 4 = − 2 2 = − 2 Since this is false, x = − 7 is an extraneous solution.
Final Answer Therefore, the only valid solution is x = − 2 .
Examples
Solving equations with square roots is a common task in physics and engineering. For example, when calculating the velocity of an object falling under gravity, you might encounter an equation involving a square root. Similarly, in electrical engineering, you might use such equations to determine the current in a circuit. Understanding how to solve these equations is crucial for making accurate predictions and designing effective systems.
