Solve: [tex] \left|x^2+2 x-5\right|=2 [/tex]
Answer (1)
Split the absolute value equation into two cases: x 2 + 2 x − 5 = 2 and x 2 + 2 x − 5 = − 2 .
Solve the first case x 2 + 2 x − 5 = 2 which simplifies to x 2 + 2 x − 7 = 0 using the quadratic formula to get x = − 1 ± 2 2 .
Solve the second case x 2 + 2 x − 5 = − 2 which simplifies to x 2 + 2 x − 3 = 0 by factoring to get ( x + 3 ) ( x − 1 ) = 0 , so x = − 3 or x = 1 .
The solutions are x = − 1 + 2 2 , − 1 − 2 2 , − 3 , 1 , so the final answer is − 1 + 2 2 , − 1 − 2 2 , − 3 , 1 .
Explanation
Understanding the Problem We are given the equation x 2 + 2 x − 5 = 2 . Our goal is to find all values of x that satisfy this equation. Because of the absolute value, we need to consider two separate cases.
Solving Case 1 Case 1: x 2 + 2 x − 5 = 2 . We can rewrite this equation as x 2 + 2 x − 7 = 0 . This is a quadratic equation of the form a x 2 + b x + c = 0 , where a = 1 , b = 2 , and c = − 7 . We can use the quadratic formula to solve for x : x = 2 a − b ± b 2 − 4 a c Substituting the values of a , b , and c , we get: x = 2 ( 1 ) − 2 ± 2 2 − 4 ( 1 ) ( − 7 ) = 2 − 2 ± 4 + 28 = 2 − 2 ± 32 = 2 − 2 ± 4 2 = − 1 ± 2 2 So the solutions for this case are x = − 1 + 2 2 and x = − 1 − 2 2 .
Solving Case 2 Case 2: x 2 + 2 x − 5 = − 2 . We can rewrite this equation as x 2 + 2 x − 3 = 0 . This quadratic equation can be factored as ( x + 3 ) ( x − 1 ) = 0 . Setting each factor equal to zero gives us the solutions x + 3 = 0 ⇒ x = − 3 and x − 1 = 0 ⇒ x = 1 .
Final Answer Combining the solutions from both cases, we have four possible values for x : x = − 1 + 2 2 , x = − 1 − 2 2 , x = − 3 , and x = 1 .
Examples
Absolute value equations are useful in many real-world scenarios, such as calculating distances or tolerances in engineering and manufacturing. For example, if you are designing a component that needs to fit within a certain tolerance, you can use absolute value equations to determine the acceptable range of dimensions. Similarly, in physics, absolute value equations can be used to model the distance an object is from a certain point, regardless of direction.
