Substitute y = x 2 to transform the equation into a quadratic equation: y 2 − 9 y + 18 = 0 .
Factor the quadratic equation: ( y − 3 ) ( y − 6 ) = 0 , which gives y = 3 or y = 6 .
Substitute back x 2 for y and solve for x : x 2 = 3 gives x = ± 3 , and x 2 = 6 gives x = ± 6 .
The real solutions are 3 , − 3 , 6 , − 6 .
Explanation
Understanding the Problem We are given the equation x 4 − 9 x 2 + 18 = 0 and asked to find all real solutions. This is a quartic equation, but we can solve it by using a substitution to turn it into a quadratic equation.
Substitution Let y = x 2 . Substituting this into the given equation, we get a quadratic equation in terms of y : y 2 − 9 y + 18 = 0
Factoring the Quadratic Equation Now, we can factor this quadratic equation: ( y − 3 ) ( y − 6 ) = 0
Solving for y Solving for y , we have two possible values: y = 3 y = 6
Substituting Back and Solving for x Now we substitute back x 2 for y to find the values of x :
If y = 3 , then x 2 = 3 , which gives us x = ± 3 .
If y = 6 , then x 2 = 6 , which gives us x = ± 6 .
Final Answer Therefore, the real solutions are x = 3 , − 3 , 6 , − 6 .
Examples
Imagine you are designing a garden and want to create a square-shaped area for planting flowers. You have a certain amount of fencing to enclose the area, and you need to determine the possible side lengths of the square. This problem is similar to finding the roots of a polynomial equation, where the roots represent the possible side lengths that satisfy the given conditions. By solving the equation, you can find the exact dimensions that meet your requirements, ensuring your garden design is both aesthetically pleasing and practical.
The real solutions of the equation x 4 − 9 x 2 + 18 = 0 are 3 , − 3 , 6 , − 6 . We used substitution to convert the quartic equation into a quadratic and then factored it to find the solutions. Each solution corresponds to values derived from the quadratic equation y 2 − 9 y + 18 = 0 .
;