Certainly! Let's go through **how to find the root(s) of a cubic equation** in detail. --- ## 1. **General Form of a Cubic Equation** A cubic equation has the form: \[ ax^3 + bx^2 + cx + d = 0 \] where \( a \neq 0 \). --- ## 2. **Step-by-Step Solution** ### **Step 1: Reduce to a Simpler Form (Depressed Cubic)** First, to simplify calculations, we **eliminate the \( x^2 \) term** by substituting: \[ x = y - \frac{b}{3a} \] Let’s compute how the equation transforms: - Expand \( x = y - \frac{b}{3a} \) - Substitute into the original equation, you get a new equation in \( y \): \[ y^3 + py + q = 0 \] where: \[ p = \frac{3ac-b^2}{3a^2} \] \[ q = \frac{2b^3-9abc+27a^2d}{27a^3} \] --- ### **Step 2: Solve the Depressed Cubic** Now, we solve: \[ y^3 + py + q = 0 \] This can be done with **Cardano's formula**. #### **Cardano's Formula** Let: \[ \Delta = \left(\frac{q}{2}\right)^2 + \left(\frac{p}{3}\right)^3 \] - If \( \Delta > 0 \): One real root, two complex roots. - If \( \Delta = 0 \): All roots real, at least two are equal. - If \( \Delta < 0 \): All roots real and unequal. Now, define: \[ A = \sqrt[3]{-\frac{q}{2} + \sqrt{\Delta}} \] \[ B = \sqrt[3]{-\frac{q}{2} - \sqrt{\Delta}} \] The **real root** is: \[ y_1 = A + B \] --- ### **Step 3: Back-Substitute for \( x \)** Recall \( x = y - \frac{b}{3a} \): \[ \boxed{ x_1 = y_1 - \frac{b}{3a} } \] This is the **real root**. --- ## 3. **Example** Let’s solve: \[ x^3 + 6x^2 + 9x + 4 = 0 \] #### **Step 1: Convert to depressed cubic** Here, \( a=1, b=6, c=9, d=4 \). \[ x = y - \frac{6}{3(1)} = y - 2 \] Calculate \( p \) and \( q \): \[ p = \frac{3*1*9 - 6^2}{3*1^2} = \frac{27 - 36}{3} = \frac{-9}{3} = -3 \] \[ q = \frac{2*6^3 - 9*1*6*9 + 27*1^2*4}{27*1^3} \] \[ 6^3 = 216,\ 2*216 = 432 \] \[ 9*1*6*9 = 486 \] \[ 27*1^2*4 = 108 \] \[ q = \frac{432 - 486 + 108}{27} = \frac{54}{27} = 2 \] So the depressed cubic is: \[ y^3 - 3y + 2 = 0 \] #### **Step 2: Cardano's Formula** \[ \Delta = \left(\frac{q}{2}\right)^2 + \left(\frac{p}{3}\right)^3 \] \[ \frac{q}{2} = 1,\ \frac{p}{3} = -1 \] \[ \Delta = (1)^2 + (-1)^3 = 1 + (-1) = 0 \] Since \( \Delta = 0 \), all roots are real, at least two are equal. \[ A = \sqrt[3]{-1 + 0} = -1 \] \[ B = \sqrt[3]{-1 - 0} = -1 \] \[ y_1 = A + B = -1 + (-1) = -2 \] \[ x_1 = y_1 - \frac{b}{3a} = -2 - 2 = -4 \] **Check for other roots:** Since \( \Delta = 0 \), we use the factor theorem or synthetic division to factor out \( (x + 4) \): Divide \( x^3 + 6x^2 + 9x + 4 \) by \( (x + 4) \): - \( x^3 + 6x^2 + 9x + 4 = (x + 4)(x^2 + 2x + 1) \) - \( x^2 + 2x + 1 = (x + 1)^2 \) So, the roots are \( x = -4, -1, -1 \). --- ## 4. **Summary Algorithm** 1. **Convert to depressed cubic**: \( x = y - \frac{b}{3a} \), get \( y^3 + py + q = 0 \). 2. **Calculate**: \( p, q, \Delta = (\frac{q}{2})^2 + (\frac{p}{3})^3 \) 3. **Apply Cardano's formula**: - If \( \Delta > 0 \): One real root by \( y_1 = A + B \) - If \( \Delta = 0 \): All roots real, at least two equal - If \( \Delta < 0 \): All roots real (use trigonometric solution) 4. **Back-substitute** to get \( x \). --- ## 5. **Special Note** - **If you find one real root (possibly by trial or rational root theorem), factor out and reduce to a quadratic.** - **Quadratic roots** can be found with the quadratic formula. --- ### **If you have a specific cubic equation, let me know, and I can solve it step by step!**