Make An Appropriate Substitution And Solve The Equation.$\left(a^2-4\right)^2+3\left(a^2-4\right)-40=0$The Solution Set Is \[$\{\square\}\$\].

Introduction

Quadratic equations are a fundamental concept in mathematics, and solving them is a crucial skill for students and professionals alike. In this article, we will focus on solving a specific quadratic equation, $\left(a^2-4\right)^2+3\left(a^2-4\right)-40=0$. We will break down the solution process into manageable steps, making it easy to understand and follow.

Understanding the Equation

The given equation is a quadratic equation in the form of $ax^2+bx+c=0$, where $a$, $b$, and $c$ are constants. In this case, the equation is $\left(a^2-4\right)^2+3\left(a^2-4\right)-40=0$. To solve this equation, we need to isolate the variable $a$.

Substitution Method

One way to solve this equation is by using the substitution method. We can substitute $x=a^2-4$ into the equation, which will simplify the equation and make it easier to solve.

Simplifying the Equation

Let's substitute $x=a^2-4$ into the equation:

$\left(a^2-4\right)^2+3\left(a^2-4\right)-40=0$

Expanding the squared term, we get:

$(a^2-4)^2 = a^4 - 8a^2 + 16$

Substituting this back into the original equation, we get:

$a^4 - 8a^2 + 16 + 3(a^2-4) - 40 = 0$

Simplifying the equation further, we get:

$a^4 - 8a^2 + 16 + 3a^2 - 12 - 40 = 0$

Combine like terms:

$a^4 - 5a^2 - 36 = 0$

Factoring the Equation

Now that we have simplified the equation, we can try to factor it. Factoring the equation will help us find the values of $a$ that satisfy the equation.

Factoring by Grouping

We can factor the equation by grouping:

$a^4 - 5a^2 - 36 = (a^4 - 36) - 5a^2$

Factoring the difference of squares, we get:

$(a^2-6)(a^2+6) - 5a^2$

Expanding the first term, we get:

$(a^2-6)(a^2+6) = a^4 + 6a^2 - 6a^2 - 36$

Combine like terms:

$a^4 - 36 - 5a^2$

Solving for $a$

Now that we have factored the equation, we can set each factor equal to zero and solve for $a$.

Solving the First Factor

Setting the first factor equal to zero, we get:

$a^2-6=0$

Adding 6 to both sides, we get:

$a^2=6$

Taking the square root of both sides, we get:

$a=\pm\sqrt{6}$

Solving the Second Factor

Setting the second factor equal to zero, we get:

$a^2+6=0$

Subtracting 6 from both sides, we get:

$a^2=-6$

Since the square of any real number cannot be negative, this equation has no real solutions.

Conclusion

In this article, we solved the quadratic equation $\left(a^2-4\right)^2+3\left(a^2-4\right)-40=0$ using the substitution method and factoring. We found that the solution set is $\{\pm\sqrt{6}\}$. This equation is a great example of how to use substitution and factoring to solve quadratic equations.

Final Answer

The final answer is $\boxed{\{\pm\sqrt{6}\}}$.

Introduction

Quadratic equations can be a challenging topic for many students and professionals. In our previous article, we solved a quadratic equation using the substitution method and factoring. In this article, we will answer some frequently asked questions about quadratic equations.

Q: What is a quadratic equation?

A: A quadratic equation is a polynomial equation of degree two, which means the highest power of the variable is two. It is typically written in the form of $ax^2+bx+c=0$, where $a$, $b$, and $c$ are constants.

Q: How do I solve a quadratic equation?

A: There are several methods to solve a quadratic equation, including factoring, the quadratic formula, and the substitution method. The method you choose will depend on the specific equation and your personal preference.

Q: What is the quadratic formula?

A: The quadratic formula is a formula that can be used to solve any quadratic equation. It is written as:

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

Q: When should I use the quadratic formula?

A: You should use the quadratic formula when the equation cannot be factored easily or when you are dealing with a complex equation.

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

A: A linear equation is a polynomial equation of degree one, which means the highest power of the variable is one. A quadratic equation, on the other hand, is a polynomial equation of degree two.

Q: Can a quadratic equation have more than two solutions?

A: Yes, a quadratic equation can have more than two solutions. However, in most cases, a quadratic equation will have two distinct solutions.

Q: How do I determine the number of solutions a quadratic equation has?

A: You can determine the number of solutions a quadratic equation has by looking at the discriminant, which is the expression under the square root in the quadratic formula. If the discriminant is positive, the equation has two distinct solutions. If the discriminant is zero, the equation has one repeated solution. If the discriminant is negative, the equation has no real solutions.

Q: What is the discriminant?

A: The discriminant is the expression under the square root in the quadratic formula. It is written as $b^2-4ac$.

Q: Can a quadratic equation have complex solutions?

A: Yes, a quadratic equation can have complex solutions. This occurs when the discriminant is negative.

Q: How do I find the complex solutions of a quadratic equation?

A: You can find the complex solutions of a quadratic equation by using the quadratic formula and taking the square root of the negative discriminant.

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

A: A polynomial equation is a general term that refers to any equation that can be written in the form of $a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0=0$, where $a_n\neq 0$. A quadratic equation, on the other hand, is a specific type of polynomial equation where the highest power of the variable is two.

Conclusion

In this article, we answered some frequently asked questions about quadratic equations. We covered topics such as the quadratic formula, the discriminant, and complex solutions. We hope that this article has been helpful in clarifying any confusion you may have had about quadratic equations.

Final Answer

The final answer is $\boxed{0}$.