Finding Real Solutions: x⁴ - 5x² + 6 Explained

When solving fourth-degree equations, the first step is to recognize patterns that can simplify our work.In this equation, notice how all terms are expressed in x squared.To simplify this equation, we can make a clever substitution. Let's replace x squared with the variable u.This transformation converts our fourth-degree equation into a much simpler quadratic equation in terms of u.This transformation is a key technique for solving certain types of higher-degree equations.Recognizing these patterns is crucial for identifying when we can use this simplification method.Let's review the key aspects of this transformation that make it so useful.Now that we understand the structure, we can proceed to solve this quadratic equation.Now that we have our quadratic equation in terms of u, let's solve it using factoring.To factor this quadratic, we need to find two numbers that multiply to give us 6 and add to give us negative 5.These numbers are negative 2 and negative 3, giving us the factored form: u minus 2 times u minus 3 equals zero.Now we can solve this equation by setting each factor equal to zero.This gives us two equations: u minus 2 equals zero, and u minus 3 equals zero.Solving these equations, we get u equals 2 or u equals 3.Now, it's crucial to remember that u represents x squared.Therefore, when we substitute back, we get x squared equals 2 or x squared equals 3.These equations will be our starting point for finding the actual values of x in the next step.Now that we have our quadratic solutions, let's find all values of x.From our quadratic solution, we found that x squared equals 2 or x squared equals 3.Let's solve x squared equals 2 first. Taking the square root of both sides, we get x equals plus or minus the square root of 2.Similarly, for x squared equals 3, we get x equals plus or minus the square root of 3.Let's verify one of our solutions by plugging it back into the original equation.Let's verify that square root of 2 is indeed a solution. Watch how the equation equals zero when we substitute this value.In conclusion, we've found all four real solutions to our fourth-degree equation.Thanks for learning about solving fourth-degree equations with Spark.E!