Problem 44 - College Algebra

Question

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

Step-by-Step Solution

Answer

The solutions to the equation \(4x^4 - 13x^2 - 9 = 0\) are \(x = ±1.8356\)

Step 1: Identify the Substitution

Since we have the equation in the form of \(4x^4 - 13x^2 - 9 = 0\), it can be noticed that if \(y\) is substituted for \(x^2\), the equation would reduce to a quadratic equation of the form \(4y^2 - 13y - 9 = 0\)

Step 2: Solve the Quadratic Equation Using the Quadratic Formula

The quadratic formula is given by \(y = (-b ± \sqrt{b^2 - 4ac}) / (2a)\). Substituting \(a = 4\), \(b = -13\), and \(c = -9\) into this, we get two possible solutions for \(y\): \(y = (13 ± \sqrt{(-13)^2 - 4*4*(-9)}) / (2*4) = (13 ± \sqrt{169 + 144}) / 8 = (13 ± \sqrt{313}) / 8 = (13 ± 2\sqrt{196})/8 = 1.625 ± 1.75\). The two solutions thus are: \(y1 = 3.375\) and \(y2 = -0.125\)

Step 3: Substitute Back For x

Since we initially substituted \(y\) for \(x^2\), at this stage, we need to find the square roots of the values of \(y\) obtained to get \(x\). Remembering that the square root of a number \(a\) gives two solutions, \(± \sqrt{a}\), we get: if \(y1 = 3.375,\) then \(x = ± \sqrt{3.375}\), and if \(y2 = -0.125\), then \(x\), being the square root of a negative number, is undefined, hence we discard this solution. Consequently, the solutions to the original quartic equation are \(x = ± \sqrt{3.375}\)

Step 4: Simplify Answers

Evaluating the values gives \(x = ±1.8356\) (rounded to four decimal places)

Key Concepts

Quadratic Equation

A quadratic equation is a type of polynomial equation that takes the form \(ax^2 + bx + c = 0\). Here, \(a\), \(b\), and \(c\) represent constants, with \(a\) not equal to zero. The term \(x^2\) signifies that this is a quadratic equation, commonly illustrated by the shape of a parabola when graphed. Understanding this structure is crucial as it allows the use of specific methods, such as the quadratic formula or factoring, to solve the equation.
A quadratic equation can have up to two real solutions since the degree of the polynomial (2, meaning the highest power of \(x\)) dictates the number of possible solutions.

• When the solutions are real and distinct, the parabola intersects the x-axis at two points.
• If there's one real, repeated solution, the curve just touches the x-axis at one point.
• With no real solutions, the parabola does not touch the x-axis at all.

In the context of the exercise, after substituting and simplifying, a quadratic equation in terms of \(y\) emerges, highlighting the importance of recognizing and solving quadratic equations.

Quartic Equation

A quartic equation is a polynomial equation of degree four, taking the general form \(ax^4 + bx^3 + cx^2 + dx + e = 0\). These types of equations can be more challenging to solve due to their higher degree, but strategies like substitution can simplify them.
In the exercise, the quartic equation \(4x^4 - 13x^2 - 9 = 0\) appears more complex than a quadratic. However, using a strategic substitution \(y = x^2\), it transforms into a simpler quadratic form. Such transformations are common when dealing with quartic equations:

• They allow for simplification by converting a higher degree into a more manageable form.
• This method relies on recognizing patterns or structures within the equation that resemble lower-degree polynomials.

Ultimately, handling quartic equations often involves breaking down the problem into smaller, solvable parts, like quadratic equations.

Square Root

The square root of a number \(a\) is a value that, when multiplied by itself, equals \(a\). Mathematically, it is expressed as \(\pm \sqrt{a}\), indicating both a positive and negative solution when dealing with real numbers.
This concept is especially pivotal in solving equations where variables are squared, as seen in the exercise's final steps:

• Upon finding a solution for \(y\), which was a stand-in for \(x^2\), determining \(x\) necessitated taking the square root of \(y\).
• This step underscores that each positive value of \(y\) yields two potential solutions for \(x\), given by \(\pm \sqrt{y}\).

However, if \(y\) is negative, \(x\) becomes complex (or imaginary) and is typically discarded if only real solutions are considered. Calculating square roots is fundamental, as it virtually "undoes" the squaring that occurred in an equation.

Quadratic Formula

The quadratic formula is a powerful tool used to solve any quadratic equation of the form \(ax^2 + bx + c = 0\). Expressed as \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), it provides solutions for \(x\) directly. This formula is derived from completing the square of a quadratic equation.
Its main advantage is its universality; the formula works for any quadratic equation, regardless of how complicated the coefficients might be:

• It calculates the solution by plugging in the coefficients \(a\), \(b\), and \(c\) from the equation.
• The discriminant \(\sqrt{b^2 - 4ac}\) is central to understanding the nature of the solutions.

In the original exercise, after substituting \(y\) for \(x^2\), the quadratic formula effectively found solutions for \(y\), which were then used to derive possible values of \(x\). The quadratic formula is a cornerstone method in algebra for solving quadratic equations, showcasing the interconnectedness of math concepts.