Solve: [tex]x=4+(4 x-4)^{\frac{1}{2}}[/tex]
Answer (2)
To solve the equation x = 4 + ( 4 x − 4 ) 2 1 , we isolated the square root, squared both sides, and simplified to form a quadratic equation. The quadratic was factored to find potential solutions, checked for validity, and the only solution is x = 10 . So, the final answer is x = 10 .
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Isolate the square root term: x − 4 = ( 4 x − 4 ) 2 1 .
Square both sides: ( x − 4 ) 2 = 4 x − 4 .
Expand and rearrange into a quadratic equation: x 2 − 12 x + 20 = 0 .
Solve the quadratic equation by factoring: ( x − 2 ) ( x − 10 ) = 0 , yielding potential solutions x = 2 and x = 10 . Check for extraneous roots. The final answer is 10 .
Explanation
Understanding the Problem We are given the equation x = 4 + ( 4 x − 4 ) 2 1 and asked to solve for x . This equation involves a square root, so we'll need to isolate the square root and square both sides to eliminate it. However, squaring both sides can introduce extraneous solutions, so we must check our final answers in the original equation.
Isolating the Square Root First, isolate the square root term by subtracting 4 from both sides of the equation: x − 4 = ( 4 x − 4 ) 2 1
Squaring Both Sides Next, square both sides of the equation to eliminate the square root: ( x − 4 ) 2 = ( 4 x − 4 )
Expanding the Equation Expand the left side of the equation: x 2 − 8 x + 16 = 4 x − 4
Rearranging into Quadratic Form Rearrange the equation into a standard quadratic equation by moving all terms to one side: x 2 − 8 x + 16 − 4 x + 4 = 0 x 2 − 12 x + 20 = 0
Solving the Quadratic Equation Now, solve the quadratic equation. We can factor the quadratic: ( x − 2 ) ( x − 10 ) = 0 This gives us two possible solutions: x = 2 and x = 10 .
Checking for Extraneous Roots We must check both solutions in the original equation to eliminate any extraneous roots. For x = 2 : 2 = 4 + ( 4 ( 2 ) − 4 ) 2 1 2 = 4 + ( 8 − 4 ) 2 1 2 = 4 + ( 4 ) 2 1 2 = 4 + 2 2 = 6 . This is false, so x = 2 is an extraneous root. For x = 10 : 10 = 4 + ( 4 ( 10 ) − 4 ) 2 1 10 = 4 + ( 40 − 4 ) 2 1 10 = 4 + ( 36 ) 2 1 10 = 4 + 6 10 = 10 . This is true, so x = 10 is a valid solution.
Final Answer Therefore, the only real solution to the equation is x = 10 .
Examples
When designing structures or calculating trajectories, engineers often encounter equations involving square roots. Solving these equations accurately is crucial for ensuring the stability and safety of designs. For instance, determining the correct tension in a cable or the precise angle of a projectile might involve solving an equation similar to the one we just tackled. The ability to manipulate and solve such equations ensures that the final design meets the required specifications and performs as expected in real-world conditions.
