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: Solve \(x^4 + x^2 - 6 = 0\)
- 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:
- \(x = \sqrt{2}\): \((\sqrt{2})^4 + (\sqrt{2})^2 - 6 = 4 + 2 - 6 = 0\)
- \(x = -\sqrt{2}\): \((- \sqrt{2})^4 + (- \sqrt{2})^2 - 6 = 4 + 2 - 6 = 0\)
Conclusion: The real solutions are \(x = \sqrt{2}\) and \(x = -\sqrt{2}\).
Example 2: Solve \(2x + 3 \sqrt{x} = 5\)
- 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 \(x = 1\).
Practice Problems with Solutions
Problem 1: Solve \(x^{2/3} - 2x^{1/3} - 8 = 0\)
- Given: \[ x^{2/3} - 2x^{1/3} - 8 = 0 \]
- Substitution: Let \(u = x^{1/3}\). Then \(u^2 = x^{2/3}\), and the equation becomes: \[ u^2 - 2u - 8 = 0 \]
- Factorization: \[ (u - 4)(u + 2) = 0 \]
- Solve for \(u\): \[ u = 4 \quad \text{or} \quad u = -2 \]
- Back-substitute \(u = x^{1/3}\): \[ x^{1/3} = 4 \implies x = 4^3 = 64 \] \[ x^{1/3} = -2 \implies x = (-2)^3 = -8 \]
Conclusion: The solutions are \(x = 64\) and \(x = -8\).
Problem 2: Solve \(x^{-2} - 3x^{-1} - 10 = 0\)
- Given: \[ x^{-2} - 3x^{-1} - 10 = 0 \]
- Substitution: Let \(u = x^{-1}\). Then \(u^2 = x^{-2}\), transforming the equation into: \[ u^2 - 3u - 10 = 0 \]
- Factorization: \[ (u - 5)(u + 2) = 0 \]
- Solve for \(u\): \[ u = 5 \quad \text{or} \quad u = -2 \]
- Back-substitute \(u = x^{-1}\): \[ \frac{1}{x} = 5 \implies x = \frac{1}{5} \] \[ \frac{1}{x} = -2 \implies x = -\frac{1}{2} \]
Conclusion: The solutions are \(x = \frac{1}{5}\) and \(x = -\frac{1}{2}\).
Problem 3 (Challenging): Solve \((x^2 - x)^2 - 8(x^2 - x) + 12 = 0\)
- Given: \[ (x^2 - x)^2 - 8(x^2 - x) + 12 = 0 \]
- Substitution: Let \(u = x^2 - x\). The equation becomes a standard quadratic in terms of \(u\): \[ u^2 - 8u + 12 = 0 \]
- Factorization: \[ (u - 6)(u - 2) = 0 \]
- Solve for \(u\): \[ u = 6 \quad \text{or} \quad u = 2 \]
- Back-substitute and solve for \(x\):
- Case 1: \(x^2 - x = 6 \implies x^2 - x - 6 = 0 \implies (x - 3)(x + 2) = 0 \implies x = 3 \text{ or } x = -2\)
- Case 2: \(x^2 - x = 2 \implies x^2 - x - 2 = 0 \implies (x - 2)(x + 1) = 0 \implies x = 2 \text{ or } x = -1\)
Conclusion: The real solutions are \(x = 3\), \(x = -2\), \(x = 2\), and \(x = -1\).