B) Hence, Solve The Equation $x^3 + X^2 - 6x = 0$.Note: Please Write Your Answer In The Format $x = \quad$

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Introduction

Solving cubic equations is a fundamental concept in algebra, and it requires a deep understanding of various mathematical techniques. In this article, we will focus on solving the cubic equation $x^3 + x^2 - 6x = 0$. This equation is a classic example of a cubic equation, and it can be solved using various methods, including factoring, synthetic division, and the rational root theorem.

Factoring the Equation

To solve the equation $x^3 + x^2 - 6x = 0$, we can start by factoring out the common term $x$. This gives us:

$$x(x^2 + x - 6) = 0$$

Now, we can see that the equation is factored into two parts: $x$ and $(x^2 + x - 6)$. We can set each part equal to zero and solve for $x$.

Setting $x = 0$

If we set $x = 0$, we get:

$$0(x^2 + x - 6) = 0$$

This equation is true for all values of $x$, so we have found one solution: $x = 0$.

Setting $(x^2 + x - 6) = 0$

Now, we need to solve the quadratic equation $(x^2 + x - 6) = 0$. We can use the quadratic formula to solve this equation:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

In this case, $a = 1$, $b = 1$, and $c = -6$. Plugging these values into the quadratic formula, we get:

$$x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-6)}}{2(1)}$$

Simplifying this expression, we get:

$$x = \frac{-1 \pm \sqrt{25}}{2}$$

$$x = \frac{-1 \pm 5}{2}$$

This gives us two possible solutions: $x = \frac{-1 + 5}{2}$ and $x = \frac{-1 - 5}{2}$.

Solving for $x$

Now, we can solve for $x$ in each of the two possible solutions:

$$x = \frac{-1 + 5}{2} = \frac{4}{2} = 2$$

$$x = \frac{-1 - 5}{2} = \frac{-6}{2} = -3$$

So, we have found two more solutions: $x = 2$ and $x = -3$.

Conclusion

In conclusion, we have solved the cubic equation $x^3 + x^2 - 6x = 0$ and found three solutions: $x = 0$, $x = 2$, and $x = -3$. These solutions can be verified by plugging them back into the original equation.

Final Answer

The final answer is: $x = 0, 2, -3$

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Introduction

In our previous article, we solved the cubic equation $x^3 + x^2 - 6x = 0$ and found three solutions: $x = 0$, $x = 2$, and $x = -3$. In this article, we will answer some frequently asked questions about solving cubic equations and provide additional insights into the solution process.

Q&A

Q: What is a cubic equation?

A: A cubic equation is a polynomial equation of degree three, which means that the highest power of the variable (in this case, $x$) is three. Cubic equations are often written in the form $ax^3 + bx^2 + cx + d = 0$, where $a$, $b$, $c$, and $d$ are constants.

Q: How do I know if a cubic equation can be factored?

A: Not all cubic equations can be factored. However, if the equation can be factored, it is often possible to find a solution by setting each factor equal to zero and solving for $x$. In the case of the equation $x^3 + x^2 - 6x = 0$, we were able to factor out the common term $x$ and solve for $x$.

Q: What is the quadratic formula?

A: The quadratic formula is a mathematical formula that can be used to solve quadratic equations of the form $ax^2 + bx + c = 0$. The quadratic formula is given by:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

In the case of the equation $(x^2 + x - 6) = 0$, we used the quadratic formula to solve for $x$.

Q: How do I know if a quadratic equation has real solutions?

A: A quadratic equation has real solutions if the discriminant ($b^2 - 4ac$) is non-negative. In the case of the equation $(x^2 + x - 6) = 0$, the discriminant is $1^2 - 4(1)(-6) = 25$, which is positive. Therefore, the equation has real solutions.

Q: What is the difference between a cubic equation and a quadratic equation?

A: A cubic equation is a polynomial equation of degree three, while a quadratic equation is a polynomial equation of degree two. Cubic equations are often more difficult to solve than quadratic equations, and may require the use of advanced mathematical techniques.

Q: Can I use the rational root theorem to solve a cubic equation?

A: Yes, the rational root theorem can be used to solve a cubic equation. However, the rational root theorem is typically used to find rational solutions to a polynomial equation, and may not be effective for finding all solutions to a cubic equation.

Q: How do I know if a cubic equation has real solutions?

A: A cubic equation has real solutions if the equation has at least one real root. In the case of the equation $x^3 + x^2 - 6x = 0$, we found three real solutions: $x = 0$, $x = 2$, and $x = -3$.

Conclusion

In conclusion, solving cubic equations can be a challenging but rewarding experience. By understanding the properties of cubic equations and using advanced mathematical techniques, it is possible to find solutions to even the most difficult equations. We hope that this article has provided you with a better understanding of how to solve cubic equations and has answered some of your frequently asked questions.

Final Answer

The final answer is: $x = 0, 2, -3$