Solve $x-5 \sqrt{x}+4=0$
Answer (1)
Substitute y = x to transform the equation into a quadratic equation y 2 − 5 y + 4 = 0 .
Factor the quadratic equation to get ( y − 1 ) ( y − 4 ) = 0 , which gives solutions y = 1 and y = 4 .
Substitute back to find x : x = 1 ⇒ x = 1 and x = 4 ⇒ x = 16 .
Verify both solutions in the original equation, confirming that x = 1 and x = 16 are valid. The final answer is 1 , 16 .
Explanation
Problem Analysis We are given the equation x − 5 x + 4 = 0 . Our goal is to find all possible values of x that satisfy this equation.
Making a Substitution Let's make a substitution to simplify the equation. Let y = x . Then, y 2 = x . Substituting these into the original equation, we get:
y 2 − 5 y + 4 = 0
Factoring the Quadratic Now we have a quadratic equation in terms of y . We can factor this equation as follows:
( y − 1 ) ( y − 4 ) = 0
Solving for y Setting each factor equal to zero, we find the solutions for y :
y − 1 = 0 ⇒ y = 1
y − 4 = 0 ⇒ y = 4
Solving for x Now we need to find the corresponding values of x . Since y = x , we have:
x = 1 ⇒ x = 1 2 = 1
x = 4 ⇒ x = 4 2 = 16
Checking the Solutions We should check our solutions in the original equation to make sure they are valid.
For x = 1 :
1 − 5 1 + 4 = 1 − 5 ( 1 ) + 4 = 1 − 5 + 4 = 0 . This solution is valid.
For x = 16 :
16 − 5 16 + 4 = 16 − 5 ( 4 ) + 4 = 16 − 20 + 4 = 0 . This solution is also valid.
Final Answer Therefore, the solutions to the equation x − 5 x + 4 = 0 are x = 1 and x = 16 .
Examples
Imagine you are designing a garden and want to calculate the area needed for a square patch of flowers. If the length of one side is related to the area by the equation x − 5 x + 4 = 0 , where x represents the area, solving this equation helps you determine the possible areas for the flower patch. Understanding how to solve such equations allows you to optimize the garden layout and ensure you have the right amount of space for your plants. This type of problem also appears in physics, where you might be analyzing the motion of an object under certain conditions, and the solutions represent specific physical parameters.
