Solve { (t-3)^2=6$}$.The Arrow Is At A Height Of 48 Ft After Approximately { \square$}$ Seconds And After { \square$}$ Seconds.

Mar 13, 2025 by ADMIN 128 views

$**Solving Quadratic Equations: A Step-by-Step Guide to ${$(t-3)^2=6\$}$**$

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 the quadratic equation ${$ (t-3)^2=6$}$, which represents the height of an object at a given time. We will break down the solution into manageable steps, providing a clear and concise explanation of each step.

Understanding the Equation

The given equation is ${$ (t-3)^2=6$}$. This equation represents a quadratic function, where ${$ t$}$ is the independent variable, and ${$ (t-3)^2$}$ is the dependent variable. The equation is in the form of ${$ ax^2+bx+c=0$}$, where ${$ a=1$}$, ${$ b=0$}$, and ${$ c=-6$}$.

Step 1: Expand the Equation

To solve the equation, we need to expand the squared term. We can do this by using the formula ${$ (x-y)^2=x2-2xy+y^2$}$. In this case, ${$ x=t$}$ and ${$ y=3$}$. Therefore, we can expand the equation as follows:

${$ (t-3)^2=t2-2(3)t+3^2$}$

Simplifying the equation, we get:

${$ t^2-6t+9=6$}$

Step 2: Rearrange the Equation

Now that we have expanded the equation, we need to rearrange it to get all the terms on one side. We can do this by subtracting ${ $6\$}$ from both sides of the equation:

${$ t^2-6t+9-6=0$}$

Simplifying the equation, we get:

${$ t^2-6t+3=0$}$

Step 3: Factor the Equation

The next step is to factor the quadratic equation. We can do this by finding two numbers whose product is ${ $3\$}$ and whose sum is ${$ -6$}$. These numbers are ${$ -3$}$ and ${$ -1$}$. Therefore, we can factor the equation as follows:

${$ (t-3)(t-1)=0$}$

Step 4: Solve for t

Now that we have factored the equation, we can solve for ${$ t$}$. We can do this by setting each factor equal to zero and solving for ${$ t$}$:

${$ t-3=0\Rightarrow t=3$}$

${$ t-1=0\Rightarrow t=1$}$

Conclusion

In this article, we have solved the quadratic equation ${$ (t-3)^2=6$}$ using the steps outlined above. We expanded the equation, rearranged it, factored it, and finally solved for ${$ t$}$. The solutions to the equation are ${$ t=3$}$ and ${$ t=1$}$. These values represent the time at which the object is at a height of ${ $48\$}$ ft.

The Height of the Object

To find the height of the object at a given time, we can substitute the value of ${$ t$}$ into the original equation. Let's say we want to find the height of the object at ${$ t=1$}$ seconds. We can substitute this value into the equation as follows:

${$ (1-3)^2=6$}$

Simplifying the equation, we get:

${$ (-2)^2=6$}$

${ $4=6\$}$

This means that the object is at a height of ${ $48\$}$ ft after approximately ${ $1\$}$ second.

The Time of the Object

To find the time at which the object is at a height of ${ $48\$}$ ft, we can substitute the value of the height into the original equation. Let's say we want to find the time at which the object is at a height of ${ $48\$}$ ft. We can substitute this value into the equation as follows:

${$ (t-3)^2=48$}$

Simplifying the equation, we get:

${$ (t-3)^2=6(8)$}$

${$ (t-3)^2=48$}$

Taking the square root of both sides of the equation, we get:

${$ t-3=\pm\sqrt{48}$}$

Simplifying the equation, we get:

${$ t-3=\pm4\sqrt{3}$}$

Adding ${ $3\$}$ to both sides of the equation, we get:

${$ t=3\pm4\sqrt{3}$}$

Therefore, the object is at a height of ${ $48\$}$ ft after approximately ${ $3+4\sqrt{3}\$}$ seconds and after approximately ${ $3-4\sqrt{3}\$}$ seconds.

Conclusion

Introduction

In our previous article, we solved the quadratic equation ${$ (t-3)^2=6$}$ using the steps outlined above. We expanded the equation, rearranged it, factored it, and finally solved for ${$ t$}$. The solutions to the equation are ${$ t=3$}$ and ${$ t=1$}$. These values represent the time at which the object is at a height of ${ $48\$}$ ft. In this article, we will answer some of the most frequently asked questions about solving quadratic equations.

Q&A

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, and ${$ x$}$ is the variable.

Q: How do I solve a quadratic equation?

A: To solve a quadratic equation, you can use the following steps:

Expand the equation, if necessary.
Rearrange the equation to get all the terms on one side.
Factor the equation, if possible.
Solve for the variable.

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

A: A quadratic equation is a polynomial equation of degree two, while a linear equation is a polynomial equation of degree one. In other words, a quadratic equation has a squared term, while a linear equation does not.

Q: Can I use a calculator to solve a quadratic equation?

A: Yes, you can use a calculator to solve a quadratic equation. However, it is always a good idea to check your work by hand to make sure you understand the solution.

Q: What is the significance of the solutions to a quadratic equation?

A: The solutions to a quadratic equation represent the values of the variable that satisfy the equation. In the case of the equation ${$ (t-3)^2=6$}$, the solutions represent the time at which the object is at a height of ${ $48\$}$ ft.

Q: Can I use the quadratic formula to solve a quadratic equation?

A: Yes, you can use the quadratic formula to solve a quadratic equation. The quadratic formula is:

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

This formula can be used to find the solutions to a quadratic equation.

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

A: A quadratic equation is a polynomial equation of degree two, while 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$}$, ${$ a_{n-1}$}$, ${$ \ldots$}$, ${$ a_1$}$, and ${$ a_0$}$ are constants, and ${$ x$}$ is the variable.

Q: Can I use the quadratic formula to solve a polynomial equation?

A: No, you cannot use the quadratic formula to solve a polynomial equation. The quadratic formula is specifically designed to solve quadratic equations, and it will not work for polynomial equations of degree three or higher.

Conclusion

In this article, we have answered some of the most frequently asked questions about solving quadratic equations. We have discussed the definition of a quadratic equation, the steps to solve a quadratic equation, and the significance of the solutions to a quadratic equation. We have also discussed the difference between a quadratic equation and a linear equation, and the difference between a quadratic equation and a polynomial equation. We hope that this article has been helpful in clarifying any confusion you may have had about solving quadratic equations.