Problem 14 - Algebra and Trigonometry Real Mathematics, Real People

Question

Solving an Equation of Quadratic Type In Exercises 13-16, find all solutions of the equation algebraically. Check your solutions. $$x^{4}-5 x^{2}-36=0$$

Step-by-Step Solution

Answer

The solutions of the equation \(x^{4}-5 x^{2}-36=0\) are \(x = 3, -3\).

Step 1: Transformation of Equation

Change the original equation \(x^{4}-5 x^{2}-36=0\) into a quadratic equation by substituting \(x^{2}\) with a new variable \(y\). This gives the new equation as \(y^{2}-5y-36=0\)

Step 2: Solving the Quadratic Equation

The roots of the quadratic equation \(y^{2}-5y-36=0\) can be found using the formula \(y = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\), where \(a=1\), \(b=-5\) and \(c=-36\). After plugging these values into the formula, we get \(y = \frac{5 \pm \sqrt{25+144}}{2}\), which results in \(y = 9, -4\)

Step 3: Backsubstitution to Find x

The original variable was \(x^{2}\), so set \(y\) equal to \(x^{2}\) to find the solutions for \(x\). When \(y=9\), \(x^{2} = 9\) which gives \(x = 3, -3\) and when \(y=-4\), \(x^{2}=-4\) has no real solution as taking square root of a negative number in real number set is not possible.

Step 4: Checking the Solutions

Substitute the solutions for \(x\) back into the original equation to make sure they are correct. Upon substitution, you will find that both \(x = 3, -3\) satisfy the original equation.

Key Concepts

Quadratic Type Equations

Quadratic type equations may look intimidating at first, but they can be managed by a clever substitution that turns them into friendlier quadratic equations. Just like the exercise equation \(x^{4}-5 x^{2}-36=0\), we look for a pattern that resembles \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. For \(x^{4}\) equations, setting \(y = x^{2}\) simplifies it to a standard quadratic form. After this transformation, we can employ traditional methods to tackle the problem, including factoring or using the quadratic formula.

Understanding this tactic requires a bit of visualization and the ability to recognize the core quadratic structure within a more complicated expression. It's an extremely useful technique that can be applied to a variety of polynomial equations, making them more manageable.

Factoring Polynomials

Factoring polynomials is a crucial skill in algebra that aids in breaking down complex expressions into simpler ones. It involves finding an equivalent expression that is a product of polynomials of lower degree. In some cases, like the transformed quadratic equation \(y^{2}-5y-36=0\), factors can be found mentally. This is typically the case with polynomials that have a leading coefficient of 1.

To successfully factor a quadratic, one must find two numbers that both multiply to the constant term, \(c\), and add up to the coefficient \(b\) in the term \(bx\). However, not all quadratics are factorable using integers, and in those cases, the quadratic formula becomes a handy tool. Polynomials of higher degrees may be factored by various methods, such as synthetic division, long division, or special factorization rules like difference of squares.

Quadratic Formula

The quadratic formula, \(y = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\), is the silver bullet for solving any quadratic equation when factoring is not feasible or quickly apparent. This formula provides the roots of \(ax^2 + bx + c = 0\) through a direct calculation. In our example, plugging in \(a=1\), \(b=-5\), and \(c=-36\) determines the potential values for \(y\), which are then used to determine the original variable \(x\) values.

It is important to recognize the discriminant part of the formula, \(\sqrt{b^{2}-4ac}\), which dictates the nature of the roots. If it's positive, there are two real solutions; if zero, there is one real solution; and if negative, the solutions involve complex numbers. Regardless of its initial form, the quadratic equation can always be solved using this fail-proof method.

Complex Numbers

Complex numbers extend the real number system and are composed of a real part and an imaginary part, usually written in the form \(a + bi\), where \(i\) is the imaginary unit, defined by \(i^{2} = -1\). They arise naturally from solving quadratic equations with a negative discriminant, as seen in the backsubstitution step where \(x^{2} = -4\) does not yield a real solution. Instead, we get two complex solutions, \(x = \pm2i\).

Introducing complex numbers allows for the formulation of a complete solution set to polynomials, as every polynomial equation of degree \(n\) will have \(n\) solutions within the realm of complex numbers. This principle is known as the Fundamental Theorem of Algebra, ensuring that even when a polynomial doesn’t have real roots, it will always have solutions in the complex plane.