Problem 37 - Precalculus: Functions and Graphs

Question

Solve the equation. \(5 y^{4}-7 y^{2}+1=0\)

Step-by-Step Solution

Answer

The solutions are \( y = \pm \sqrt{\frac{7+\sqrt{29}}{10}} \) and \( y = \pm \sqrt{\frac{7-\sqrt{29}}{10}} \).

Step 1: Substitution to Simplify

Introduce a substitution to make the equation easier to solve. Let \( z = y^2 \). The equation becomes \( 5z^2 - 7z + 1 = 0 \). Now, we are dealing with a quadratic equation in terms of \( z \).

Step 2: Identify and Apply the Quadratic Formula

The quadratic equation in standard form is \( az^2 + bz + c = 0 \). Here, \( a = 5 \), \( b = -7 \), and \( c = 1 \). Use the quadratic formula \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to find the values of \( z \).

Step 3: Calculate the Discriminant

Compute the discriminant: \( b^2 - 4ac = (-7)^2 - 4 \cdot 5 \cdot 1 = 49 - 20 = 29 \). Since the discriminant is positive, there will be two real roots for \( z \).

Step 4: Solve for z Using the Quadratic Formula

Substitute the values of \( a \), \( b \), and \( c \) into the quadratic formula:\[ z = \frac{7 \pm \sqrt{29}}{10} \]This gives two solutions: \( z_1 = \frac{7 + \sqrt{29}}{10} \) and \( z_2 = \frac{7 - \sqrt{29}}{10} \).

Step 5: Reverse Substitution to Find y

Since \( z = y^2 \), solve for \( y \) by taking the square root of both solutions for \( z \). Thus, \( y = \pm \sqrt{z} \).The solutions for \( y \) are:\[ y = \pm \sqrt{\frac{7 + \sqrt{29}}{10}} \]\[ y = \pm \sqrt{\frac{7 - \sqrt{29}}{10}} \]

Step 6: Final Answer Evaluation

The four possible solutions for the equation are:\( y = \sqrt{\frac{7 + \sqrt{29}}{10}}, y = -\sqrt{\frac{7 + \sqrt{29}}{10}}, y = \sqrt{\frac{7 - \sqrt{29}}{10}}, \text{and} y = -\sqrt{\frac{7 - \sqrt{29}}{10}} \). These solutions are checked and confirmed as correct.

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). It allows you to find the roots, or solutions, of the equation by substituting the coefficients \( a \), \( b \), and \( c \) into the formula. The quadratic formula is given by:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

This formula stems from the process of completing the square and works for any quadratic equation, regardless of whether the roots are real or complex. To use the formula, follow these steps:

• Identify the coefficients \( a \), \( b \), and \( c \) in your quadratic equation.
• Substitute them into the quadratic formula.
• Simplify to find the values of \( x \), which are the solutions to the equation.

Discriminant

The discriminant plays a crucial role in determining the nature of the roots of a quadratic equation. It is the part of the quadratic formula under the square root, specifically \( b^2 - 4ac \). The discriminant helps us understand how many and what type of roots the equation has:

• If the discriminant is positive, there are two distinct real roots.
• If the discriminant is zero, there is exactly one real root, because the roots are repeated.
• If the discriminant is negative, there are no real roots, only complex ones.

In our example problem, the discriminant is calculated as 29, which is positive. This indicates that the equation has two distinct real roots. Calculating this value is a vital step before applying the quadratic formula, as it informs us about the nature of our solutions.

Substitution Method

Substitution is a problem-solving technique used to simplify equations, particularly helpful in solving quadratic equations that appear in a more complex form. By substituting a variable to reduce the equation into a standard quadratic form, we make it easier to solve. Here's how it works:

• Identify a variable and make a substitution that simplifies part of the equation. For instance, in the given equation \( 5y^4 - 7y^2 + 1 = 0 \), we let \( z = y^2 \). This changes the equation to \( 5z^2 - 7z + 1 = 0 \).
• Solve the resulting quadratic equation using appropriate methods, like the quadratic formula.
• Once you have the solutions for the substituted variable, reverse the substitution to get back to the original variable.

This approach not only simplifies the equation but also allows us to apply well-known methods like the quadratic formula to find the roots effectively.

Real Roots

Real roots are the solutions of an equation that are real numbers. In the context of quadratic equations and solving using the quadratic formula, the nature of the roots is determined by the discriminant. For our problem, since the discriminant \( b^2 - 4ac \) is positive, the equation has two real roots.

• A real root is simply a solution that belongs to the set of real numbers, implying that it has no imaginary component.
• Finding real roots involves solving the quadratic equation until you simplify it to values without any imaginary parts, ensuring they fit within the real number system.

In our specific case, solving \( 5z^2 - 7z + 1 = 0 \) provided us with real values for \( z \). Subsequently, taking the square roots (considering \( z = y^2 \)) and finding both positive and negative solutions led us to identify the four real roots for \( y \). This process underlines that having a positive discriminant is the key to identifying real, distinct roots in such equations.