Find all real solutions. (Enter your answers as comma-separated lists. If there is no real solution, enter NO REALSOLUTION.) [tex]\begin{array}{l} x^{\frac{1}{2}}-4 x^{\frac{1}{4}}+3=0 \\ x=? \end{array}[/tex]
Answer (2)
Substitute y = x 4 1 to transform the equation into a quadratic equation y 2 − 4 y + 3 = 0 .
Factor the quadratic equation to find the solutions for y : y = 1 and y = 3 .
Substitute back to find the solutions for x : x = y 4 , which gives x = 1 and x = 81 .
Verify the solutions in the original equation. The real solutions are 1 , 81 .
Explanation
Understanding the Problem We are given the equation x 2 1 − 4 x 4 1 + 3 = 0 . Our goal is to find all real solutions for x .
Making a Substitution Let's make a substitution to simplify the equation. Let y = x 4 1 . Then, y 2 = ( x 4 1 ) 2 = x 2 1 . Substituting these into the original equation, we get a quadratic equation in terms of y : y 2 − 4 y + 3 = 0
Solving the Quadratic Equation Now, we solve the quadratic equation for y . We can factor the quadratic as follows: ( y − 1 ) ( y − 3 ) = 0 This gives us two possible values for y : y = 1 or y = 3
Finding the Values of x Since y = x 4 1 , we have x = y 4 . We can now find the corresponding values of x for each value of y .If y = 1 , then x = 1 4 = 1 If y = 3 , then x = 3 4 = 81
Checking the Solutions We need to check if these solutions satisfy the original equation.For x = 1 : ( 1 ) 2 1 − 4 ( 1 ) 4 1 + 3 = 1 − 4 ( 1 ) + 3 = 1 − 4 + 3 = 0 So, x = 1 is a valid solution.For x = 81 : ( 81 ) 2 1 − 4 ( 81 ) 4 1 + 3 = 9 − 4 ( 3 ) + 3 = 9 − 12 + 3 = 0 So, x = 81 is also a valid solution.
Final Answer Therefore, the real solutions for x are 1 and 81 .
Examples
Imagine you are designing a water filtration system where the flow rate is modeled by the equation x 2 1 − 4 x 4 1 + 3 = 0 , where x represents the filter's surface area. Solving this equation helps you determine the optimal filter size to achieve a specific flow rate. Understanding how to manipulate and solve such equations is crucial in engineering to ensure systems operate efficiently and effectively. This type of problem also appears in financial modeling, where x could represent an investment amount, and the equation models the return on investment.
The real solutions to the equation x 2 1 − 4 x 4 1 + 3 = 0 are x = 1 and x = 81 . We reached these solutions by substituting y = x 4 1 , simplifying to a quadratic equation, and then finding the corresponding values of x . Both solutions were verified to satisfy the original equation.
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