Problem 35 - Algebra for College Students

Question

Solve each equation. $$3 x^{4}-35 x^{2}+72=0$$

Step-by-Step Solution

Answer

Solutions: \( x = 3, -3, \sqrt{\frac{8}{3}}, -\sqrt{\frac{8}{3}} \).

Step 1: Substitution

Start by simplifying the equation by substituting a new variable. Let \( y = x^2 \). Then, \( x^4 = y^2 \) and the equation becomes:\[ 3y^2 - 35y + 72 = 0 \] This is now a quadratic equation in terms of \( y \).

Step 2: Applying the Quadratic Formula

Solve the quadratic equation using the quadratic formula: \( y = \frac{-b \, \pm \, \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 3 \), \( b = -35 \), and \( c = 72 \). Plugging these values in, we find:\[ y = \frac{35 \, \pm \, \sqrt{(-35)^2 - 4 \times 3 \times 72}}{2 \times 3} \]\[ y = \frac{35 \, \pm \, \sqrt{1225 - 864}}{6} \]\[ y = \frac{35 \, \pm \, \sqrt{361}}{6} \]\[ y = \frac{35 \, \pm \, 19}{6} \]

Step 3: Solving for \( y \)

Calculate the two possible values for \( y \):\[ y_1 = \frac{35 + 19}{6} = \frac{54}{6} = 9 \]\[ y_2 = \frac{35 - 19}{6} = \frac{16}{6} = \frac{8}{3} \]

Step 4: Back-Substitution for \( x \)

The original substitution was \( y = x^2 \). Now solve for \( x \) from \( y_1 \) and \( y_2 \):- For \( y = 9 \): \( x^2 = 9 \), so, \( x = \pm 3 \).- For \( y = \frac{8}{3} \): \( x^2 = \frac{8}{3} \), so, \( x = \pm \, \sqrt{\frac{8}{3}} \).

Step 5: Conclusion

The solutions for \( x \) are \( x = 3 \), \( x = -3 \), \( x = \sqrt{\frac{8}{3}} \), and \( x = -\sqrt{\frac{8}{3}} \).

Key Concepts

Quadratic Formula

The quadratic formula is a vital tool for solving quadratic equations of any form. It is especially helpful when factoring the equation is difficult or impossible. The formula is defined as:\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Where:

• \(a\), \(b\), and \(c\) are coefficients from the quadratic equation \(ay^2 + by + c = 0\).
• \(-b\) and \(\pm\sqrt{b^2 - 4ac}\) determine the roots of the equation.

Each part of the quadratic formula plays a crucial role.
The term \(b^2 - 4ac\), known as the discriminant, tells us about the nature of the roots:

• If it's positive, there are two real and distinct roots.
• If it's zero, there are two real and identical roots (one solution).
• If it's negative, the roots are complex and not real numbers.

Remember, using the quadratic formula allows you to efficiently find the roots of any standard quadratic equation.

Substitution Method

The substitution method is a strategic approach used to simplify solving equations, especially higher-degree polynomials. When dealing with complex polynomials, substitution transforms them into simpler forms, often into quadratic equations.
In the given problem, we started with the equation:\[3x^4 - 35x^2 + 72 = 0\]By substituting \(y = x^2\), the polynomial equation becomes easier to solve:\[3y^2 - 35y + 72 = 0\]Why choose substitution?

• Reduces complexity by lowering the degree of polynomial, simplifying calculations.
• Makes it possible to apply methods like the quadratic formula effectively.
• Transforms the problem into a form familiar to solving quadratic equations.

This technique streamlines solving and ensures accuracy by eliminating unnecessary complexity in equations.

Roots of Equations

Roots of equations are the solutions to the equation where the output equals zero. In solving polynomial equations, the roots provide critical insights about the behavior of the equation on a graph.
For the reduced quadratic equation \(3y^2 - 35y + 72 = 0\), applying the quadratic formula yields two roots:

• \(y_1 = 9\)
• \(y_2 = \frac{8}{3}\)

These roots tell us the points where the graph intersects the y-axis. Knowing the roots allows solving for the variable in the original equation:

• For \(y = 9\), solve \(x^2 = 9\), resulting in \(x = \pm 3\).
• For \(y = \frac{8}{3}\), solve \(x^2 = \frac{8}{3}\), giving \(x = \pm \sqrt{\frac{8}{3}}\).

The roots reveal all possible solutions for \(x\). These solutions represent where the values of \(x\) make the original polynomial equation true. Identifying roots is essential in understanding how equations behave and are key when graphically representing functions.