Problem 37 - Intermediate Algebra

Question

Solve each equation. $$ 3 x^{4}+17 x^{2}+20=0 $$

Step-by-Step Solution

Answer

The equation has no real solutions.

Step 1: Recognize the Quadratic Form

The equation \(3x^4 + 17x^2 + 20 = 0\) is a bi-quadratic equation. Notice that if we let \(y = x^2\), then \(x^4\) becomes \(y^2\). This makes the equation transform into a quadratic form: \(3y^2 + 17y + 20 = 0\).

Step 2: Apply the Quadratic Formula

We use the quadratic formula to solve for \(y\) in the equation \(3y^2 + 17y + 20 = 0\), where \(a = 3\), \(b = 17\), \(c = 20\). The quadratic formula is \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

Step 3: Calculate the Discriminant

First, compute the discriminant \(b^2 - 4ac\): \[b^2 - 4ac = 17^2 - 4 \cdot 3 \cdot 20 = 289 - 240 = 49\] Since the discriminant is 49, a perfect square, we can continue using the quadratic formula.

Step 4: Solve for \(y\)

Plug the values into the quadratic formula: \[y = \frac{-17 \pm \sqrt{49}}{6} = \frac{-17 \pm 7}{6}\]. This gives us two solutions for \(y\): \(y = \frac{-10}{6} = -\frac{5}{3}\) and \(y = \frac{-24}{6} = -4\).

Step 5: Solve for \(x\) from \(y = x^2\)

Recall that \(y = x^2\). We substitute back to find \(x\): For \(y = -\frac{5}{3}\): We have no real solution since \(x^2\) can't be negative in real numbers. For \(y = -4\): Similarly, no real solution exists.

Step 6: Conclude the Solution

Since \(y = -\frac{5}{3}\) and \(y = -4\), and both don't yield real roots for \(x^2\), there are no real solutions to the original equation.

Key Concepts

Bi-quadratic Equations

Bi-quadratic equations are special types of polynomial equations. They are characterized by having the highest exponent of a variable as four, such as in the equation \(3x^4 + 17x^2 + 20 = 0\). In general, a bi-quadratic equation can be expressed in the form \(ax^4 + bx^2 + c = 0\). Notice that there is no linear term (like \(x\)), simplifying to just powers of four and two.

To solve a bi-quadratic equation, a common method is to use a substitution. Here, substituting \(y = x^2\) transforms the equation into a simpler quadratic form, \(ay^2 + by + c = 0\). This substitution is possible because \(x^4\) becomes \((x^2)^2\), that is \(y^2\).

Once the substitution is done, you deal with a quadratic equation that is easier to solve using familiar methods. After solving for \(y\), converting back to \(x\) involves checking solutions for validity, especially with respect to real or complex numbers.

Quadratic Formula

The quadratic formula is a fundamental tool in algebra for solving quadratic equations, namely equations of the form \(ax^2 + bx + c = 0\). The formula itself is \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). This formula provides the solutions for \(y\) when the quadratic equation coefficients \(a\), \(b\), and \(c\) are known.

Using the quadratic formula begins with identifying the coefficients in the equation. In our bi-quadratic equation example, we substitute \(y = x^2\) first, then tackle the equation \(3y^2 + 17y + 20 = 0\), where \(a = 3\), \(b = 17\), and \(c = 20\).

After computing the discriminant, we plug the values into the formula to find the solutions for \(y\). It’s crucial to perform each step carefully to ensure accuracy, as errors could lead to incorrect roots or overlooked complex solutions.

Discriminant

The discriminant plays a pivotal role in determining the nature of the solutions of a quadratic equation. It is given by the expression \(b^2 - 4ac\).

For every quadratic equation \(ax^2 + bx + c = 0\), evaluating the discriminant helps us understand the type and number of roots:

• If the discriminant is positive and a perfect square, the quadratic equation has two distinct real roots.
• If the discriminant is positive but not a perfect square, the roots are real but irrational.
• If the discriminant equals zero, there is exactly one real root, often called a repeated root.
• If the discriminant is negative, the roots are complex and occur as a pair of conjugates.

In our exercise, the discriminant \(b^2 - 4ac = 49\) is a perfect square (\(7^2\)), indicating two distinct real roots for \(y\). However, translating these roots to \(x\) might lead to non-real solutions, as our equation conversion shows. Always double-check the implications of the discriminant on your solutions to ensure the validity of real or complex solutions.