Question

Find the real solutions of each equation. $$ x^{4}-10 x^{2}+25=0 $$

Short answer

The solutions are \[ x = \sqrt{5} \] and \[ x = -\sqrt{5} \].

Step-by-step solution

Substitute a New Variable

Let’s introduce a new variable to simplify the equation. Let’s set \[ y = x^2 \] which transforms the original equation into \[ y^2 - 10y + 25 = 0 \] .

Solve the Quadratic Equation

Solve the quadratic equation \[ y^2 - 10y + 25 = 0 \]. This can be factored as \[ (y - 5)^2 = 0 \]. Thus, we have \[ y = 5 \].

Substitute Back to the Original Variable

Recall our substitution \[ y = x^2 \]. Now, substitute back to get \[ x^2 = 5 \].

Solve for x

Solve for \[ x \] by taking the square root of both sides: \[ x = \pm \sqrt{5} \].

Key concepts

quartic equations

A quartic equation is a type of algebraic equation where the highest degree of the variable is four, which means it has the form: \[ a x^4 + b x^3 + c x^2 + d x + e = 0 \] In this exercise, we are dealing with the quartic equation:\[ x^4 - 10 x^2 + 25 = 0 \]. Quartic equations can be more complex than quadratic or cubic equations due to the higher degree. However, they often have special methods for solving them, such as factoring or using algebraic substitutions to simplify them. Understanding how to reduce a quartic equation to a simpler form is key to finding its solutions.

quadratic substitution

Quadratic substitution is a useful method when dealing with quartic and other higher degree polynomial equations. By substituting a new variable, you can transform the equation into a more manageable quadratic form. In our example, we start with: \[ x^4 - 10 x^2 + 25 = 0 \] We introduce a new variable, letting: \[ y = x^2 \] This substitution transforms the original quartic equation into: \[ y^2 - 10 y + 25 = 0 \] Now, we have a quadratic equation which is easier to solve. Once this simpler equation is solved, we substitute back to find the solutions in terms of the original variable.

factoring

Factoring is a method used to simplify equations and solve for the roots or solutions of an equation. In this context, after applying quadratic substitution, our equation is: \[ y^2 - 10 y + 25 = 0 \] This quadratic equation can be factored as: \[ (y - 5)^2 = 0 \] Factoring reveals that: \[ y - 5 = 0 \] This simplifies to: \[ y = 5 \] Finally, remember that we originally set \[ y = x^2 \], so we substitute back to find: \[ x^2 = 5 \] To find the values of \[ x \], we take the square root of both sides, resulting in: \[ x = \pm \sqrt{5} \] This factoring method simplifies and helps in solving complex polynomial equations by reducing them to more straightforward forms.