Solve Equations of the Quadratic Form

This tutorial explains how to solve equations that can be written in quadratic form. Step-by-step examples with detailed solutions and explanations are provided.

Review

A quadratic equation has the general form:

\[ ax^2 + bx + c = 0 \]

with \(a \neq 0\). There are several methods to solve quadratic equations. In this tutorial, we use the quadratic formula and discriminants and the factoring method.

Examples with Solutions

Example 1

Find all real solutions to the equation:

\[ x^4 + x^2 - 6 = 0 \]

Solution:

• Given: \[ x^4 + x^2 - 6 = 0 \]
• Substitution: Let \(u = x^2\). Then the equation becomes: \[ u^2 + u - 6 = 0 \]
• Factorization: \[ (u + 3)(u - 2) = 0 \]
• Zero factor theorem: \[ u + 3 = 0 \quad \text{or} \quad u - 2 = 0 \]
• Solve for \(u\): \[ u = -3 \quad \text{or} \quad u = 2 \]
• Back-substitute \(u = x^2\): \[ x^2 = -3 \quad (\text{no real solution}), \quad x^2 = 2 \implies x = \pm \sqrt{2} \]

Check Solutions:

1. \(x = \sqrt{2}\): \((\sqrt{2})^4 + (\sqrt{2})^2 - 6 = 4 + 2 - 6 = 0\)
2. \(x = -\sqrt{2}\): \((- \sqrt{2})^4 + (- \sqrt{2})^2 - 6 = 4 + 2 - 6 = 0\)

Conclusion: The real solutions are \(x = \sqrt{2}\) and \(x = -\sqrt{2}\).

Matched Exercise 1: Find all real solutions to \[ x^4 - 2x^2 - 3 = 0 \]

Answer to Matched Exercise

Example 2

Find all real solutions to the equation:

\[ 2x + 3 \sqrt{x} = 5 \]

Solution:

• Given: \[ 2x + 3 \sqrt{x} = 5 \]
• Substitution: Let \(u = \sqrt{x} \ge 0\). Then \[ 2u^2 + 3u - 5 = 0 \]
• Discriminant: \[ D = b^2 - 4ac = 3^2 - 4 \cdot 2 \cdot (-5) = 49 \]
• Quadratic formula: \[ u = \frac{-b \pm \sqrt{D}}{2a} = \frac{-3 \pm 7}{4} \implies u_1 = 1, \ u_2 = -\frac{5}{2} \]
• Back-substitute: \[ \sqrt{x} = 1 \implies x = 1, \quad \sqrt{x} = -\frac{5}{2} \ (\text{no real solution}) \]

Check Solution: \(x = 1\) satisfies \(2(1) + 3\sqrt{1} = 5\).

Conclusion: The real solution is \(x = 1\).

Matched Exercise 2: Find all real solutions to \[ x - 3\sqrt{x} - 4 = 0 \]

Answer to Matched Exercise

Solutions to Matched Exercises

Matched Exercise 1

Solve \[ x^4 - 2x^2 - 3 = 0 \]

Let \(u = x^2\). Then \[ u^2 - 2u - 3 = 0 \implies u = -1 \ (\text{no real solution}), \ u = 3 \implies x = \pm \sqrt{3} \]

Matched Exercise 2

Solve \[ x - 3\sqrt{x} - 4 = 0 \]

Let \(u = \sqrt{x}\). Then \[ u^2 - 3u - 4 = 0 \implies u = -1 \ (\text{no real solution}), \ u = 4 \implies x = 16 \]

More References and Links

Solve Equations, Systems of Equations and Inequalities