Solve: [tex]s =4+\sqrt{ s +2}[/tex]
Answer (2)
To solve the equation s = 4 + s + 2 , we isolate the square root and square both sides to form a quadratic equation. After factoring and checking possible solutions, we find that the only valid solution is s = 7 .
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Isolate the square root: s + 2 = s − 4 .
Square both sides: s + 2 = s 2 − 8 s + 16 .
Rearrange into a quadratic equation: s 2 − 9 s + 14 = 0 .
Solve by factoring: ( s − 2 ) ( s − 7 ) = 0 , yielding s = 2 or s = 7 . Check for extraneous solutions. The final answer is 7 .
Explanation
Problem Analysis We are given the equation s = 4 + s + 2 and asked to solve for s . We will isolate the square root, square both sides, and then solve the resulting quadratic equation. Finally, we must check our solutions to make sure they are not extraneous.
Isolating the Square Root First, isolate the square root term by subtracting 4 from both sides of the equation: s + 2 = s − 4
Squaring Both Sides Next, square both sides of the equation to eliminate the square root: ( s + 2 ) 2 = ( s − 4 ) 2
s + 2 = s 2 − 8 s + 16
Rearranging into Quadratic Form Now, rearrange the equation into a standard quadratic equation by subtracting s and 2 from both sides: 0 = s 2 − 9 s + 14
Factoring the Quadratic We can solve this quadratic equation by factoring. We are looking for two numbers that multiply to 14 and add to -9. These numbers are -2 and -7. Thus, we can factor the quadratic as: ( s − 2 ) ( s − 7 ) = 0
Finding Possible Solutions This gives us two possible solutions for s : s = 2 or s = 7
Checking s=2 We must check these solutions in the original equation to eliminate any extraneous solutions. For s = 2 : 2 = 4 + 2 + 2
2 = 4 + 4
2 = 4 + 2
2 = 6
This is false, so s = 2 is an extraneous solution.
Checking s=7 For s = 7 : 7 = 4 + 7 + 2
7 = 4 + 9
7 = 4 + 3
7 = 7
This is true, so s = 7 is a valid solution.
Final Answer Therefore, the only real solution to the equation is s = 7 .
Examples
When designing a bridge, engineers often use equations involving square roots to calculate the tension and compression forces in the supporting cables. Solving these equations ensures the bridge's stability and safety. Similarly, in physics, the velocity of an object under constant acceleration can be modeled using equations with square roots. Finding the correct solution helps predict the object's position and motion accurately. These real-world applications highlight the importance of solving equations with square roots and verifying the solutions.
