$\begin{array}{l}
(x-2)(x-3)=0 \\
x=2 x \geqslant 3
\end{array}$
10. Find all real solutions.
$x^{\frac{1}{2}}-4 x^{\frac{1}{4}}+3=0$
Answer (1)
Substitute y = x 4 1 to transform the equation into a quadratic equation: y 2 − 4 y + 3 = 0 .
Factor the quadratic equation: ( y − 1 ) ( y − 3 ) = 0 , which gives y = 1 or y = 3 .
Solve for x using x = y 4 : If y = 1 , then x = 1 ; if y = 3 , then x = 81 .
The real solutions are 1 , 81 .
Explanation
Understanding the Problem We are given the equation x 2 1 − 4 x 4 1 + 3 = 0 and asked to find all real solutions.
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: 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 Now we need to find the corresponding values of x . Recall that we defined y = x 4 1 .
If y = 1 , then x 4 1 = 1 . Raising both sides to the power of 4, we get x = 1 4 = 1 .
If y = 3 , then x 4 1 = 3 . Raising both sides to the power of 4, we get x = 3 4 = 81 .
Checking the Solutions We should check our solutions in the original equation to make sure they are valid.
For x = 1 , we have 1 2 1 − 4 ( 1 4 1 ) + 3 = 1 − 4 ( 1 ) + 3 = 1 − 4 + 3 = 0 , so x = 1 is a solution.
For x = 81 , we have 8 1 2 1 − 4 ( 8 1 4 1 ) + 3 = 9 − 4 ( 3 ) + 3 = 9 − 12 + 3 = 0 , so x = 81 is a solution.
Final Answer Therefore, the real solutions to the equation x 2 1 − 4 x 4 1 + 3 = 0 are x = 1 and x = 81 .
Examples
Imagine you are designing a water fountain where the water jet's height is determined by the equation x 2 1 − 4 x 4 1 + 3 = 0 , where x represents the water pressure. By solving this equation, you find the specific pressure settings ( x = 1 and x = 81 ) that result in the water jet reaching a desired height of zero. This ensures the fountain operates correctly and provides the intended visual effect. Understanding how to solve such equations allows for precise control and optimization in various engineering and design applications.
