Problem 42 - College Algebra

Question

Solve each equation in by making an appropriate substitution. $$ 4 x^{4}=13 x^{2}-9 $$

Step-by-Step Solution

Answer

After applying the Quadratic Formula, we find two solutions for \(u\). Each will then lead to two solutions for \(x\), giving us potentially 4 different possibilities for \(x\). It's important to test these solutions in the original equation to check their validity, as some might be extraneous.

Step 1: Substitution

Choose \(u = x^2\). Substitute it into our equation. We get: \(4u^2=13u-9\).

Step 2: Rearrange the quadratic equation

Rearrange our quadratic equation to have 0 on one side: \(4u^2 - 13u +9 = 0\).

Step 3: Apply the Quadratic Formula

Solve for \(u\) using the Quadratic Formula. \[u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here \(a = 4\), \(b = -13\), and \(c = 9\), so we get \[u = \frac{13 \pm \sqrt{(-13)^2 - 4*4*9}}{2*4}\]. That gives us two solutions for \(u\).

Step 4: Solve for \(x\)

Remember that our substitution was \(u = x^2\) so we have to solve for \(x\) now. We get \(x = \pm \sqrt{u}\). If \(u\) has more than one solution, this will lead to multiple possible values for \(x\).

Key Concepts

Quadratic Substitution Method

The quadratic substitution method is a useful technique for solving higher-degree polynomial equations that can be transformed into quadratic form. This method involves substituting a variable, typically with the notation of 'u', to represent a part of the original equation. In the given exercise, the substitution is made with u = x^2, which simplifies the fourth-degree equation into a quadratic one. The transformed equation, 4u^2 - 13u + 9 = 0, is more manageable and can then be approached using methods suitable for quadratic equations, such as factoring, completing the square, or applying the quadratic formula.

Following a successful substitution, a crucial step is to correctly reverse the substitution process after solving the quadratic equation. This requires substituting back to the original variable and solving for the real values of x. As seen in the exercise, the results for u are re-substituted as x = \( \pm \sqrt{u} \) to find the final solutions for x.

A key takeaway from this method is ensuring that the substitution leads to a solvable quadratic equation and that the back-substitution is accurately executed to find the original variable's values.

Quadratic Formula

The quadratic formula is probably the most famous method for solving quadratic equations and is derived from the process of completing the square. It provides a straightforward solution to any quadratic equation in the standard form ax^2 + bx + c = 0. The formula is often cited as x = \(\frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\), where a, b, and c are coefficients from the quadratic equation.

To use the formula effectively, identify the values of a, b, and c from the given equation and substitute them into the formula accordingly. The term b^2 - 4ac within the square root is known as the discriminant, and it is critical as it determines the nature and quantity of the equation's roots. A positive discriminant indicates two distinct real roots, zero means one real root (or a repeated root), while a negative discriminant implies there are no real roots.

The quadratic formula is extremely versatile, providing solutions when factoring is difficult or impossible. However, simplifying the terms under the square root and rationalizing the denominator if necessary are important for finding the simplest form of the solutions.

Factoring Quadratics

Factoring quadratics is an algebraic process where we express a quadratic equation as a product of its factors. To factor a quadratic equation in the form ax^2 + bx + c = 0, we look for two binomials ((dx + e)(fx + g) = 0) that multiply to give the original quadratic equation. The product of the first terms d and f must equal a, the product of the outer and inner terms must add up to b, and the product of the last terms e and g must equal c.

When a quadratic is factorable, setting each binomial equal to zero provides the roots of the equation (dx + e = 0 and fx + g = 0). It is a quick and effective method when the equation easily lends itself to factoring, especially when a, b, and c are small integers. However, in more complicated cases or when dealing with larger coefficients, factoring may be challenging or infeasible, prompting the use of other methods such as the quadratic formula. Facilitating the understanding of the factoring process, often utilizing the Distributive Property and looking for common factors or patterns, can be crucial for students to master solving quadratic equations.

A key tip for factoring is to always look for a Greatest Common Factor (GCF) before attempting to factor quadratics, as simplifying the equation first can make the factoring process much smoother and more apparent.