What Are The Solutions To The Quadratic Equation $(5y + 6)^2 = 24$?A. $y = \frac{-6 + 2\sqrt{6}}{5}$ And $y = \frac{-6 - 2\sqrt{6}}{5}$B. $y = \frac{-6 + 2\sqrt{6}}{5}$ And $y = \frac{6 - 2\sqrt{6}}{5}$C.

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Introduction

The quadratic equation is a fundamental concept in mathematics, and it has numerous applications in various fields, including physics, engineering, and economics. The quadratic equation is a polynomial equation of degree two, which means the highest power of the variable is two. In this article, we will focus on solving the quadratic equation (5y+6)2=24(5y + 6)^2 = 24.

Understanding the Quadratic Equation

The quadratic equation is typically written in the form ax2+bx+c=0ax^2 + bx + c = 0, where aa, bb, and cc are constants, and xx is the variable. However, in this case, we have a quadratic equation in the form of (5y+6)2=24(5y + 6)^2 = 24. To solve this equation, we need to expand the left-hand side and then simplify it.

Expanding the Left-Hand Side

To expand the left-hand side of the equation, we need to use the formula (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2. In this case, a=5ya = 5y and b=6b = 6. Therefore, we can expand the left-hand side as follows:

(5y+6)2=(5y)2+2(5y)(6)+62(5y + 6)^2 = (5y)^2 + 2(5y)(6) + 6^2

Simplifying the Equation

Now that we have expanded the left-hand side, we can simplify the equation by combining like terms. We have:

(5y)2=25y2(5y)^2 = 25y^2 2(5y)(6)=60y2(5y)(6) = 60y 62=366^2 = 36

Therefore, the equation becomes:

25y2+60y+36=2425y^2 + 60y + 36 = 24

Rearranging the Equation

To solve the equation, we need to rearrange it to the standard form of a quadratic equation, which is ax2+bx+c=0ax^2 + bx + c = 0. We can do this by subtracting 24 from both sides of the equation:

25y2+60y+12=025y^2 + 60y + 12 = 0

Solving the Quadratic Equation

Now that we have the equation in the standard form, we can solve it using the quadratic formula. The quadratic formula is given by:

x=βˆ’bΒ±b2βˆ’4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

In this case, a=25a = 25, b=60b = 60, and c=12c = 12. Therefore, we can plug these values into the quadratic formula to get:

y=βˆ’60Β±602βˆ’4(25)(12)2(25)y = \frac{-60 \pm \sqrt{60^2 - 4(25)(12)}}{2(25)}

Simplifying the Quadratic Formula

To simplify the quadratic formula, we need to calculate the value of the expression inside the square root. We have:

602=360060^2 = 3600 4(25)(12)=12004(25)(12) = 1200

Therefore, the expression inside the square root becomes:

3600βˆ’1200=24003600 - 1200 = 2400

Simplifying the Quadratic Formula (continued)

Now that we have simplified the expression inside the square root, we can simplify the quadratic formula further. We have:

y=βˆ’60Β±240050y = \frac{-60 \pm \sqrt{2400}}{50}

Simplifying the Square Root

To simplify the square root, we need to find the square root of 2400. We have:

2400=48.98\sqrt{2400} = 48.98

Therefore, the quadratic formula becomes:

y=βˆ’60Β±48.9850y = \frac{-60 \pm 48.98}{50}

Simplifying the Quadratic Formula (continued)

Now that we have simplified the quadratic formula, we can simplify it further by calculating the two possible values of yy. We have:

y=βˆ’60+48.9850=βˆ’11.0250=βˆ’0.2204y = \frac{-60 + 48.98}{50} = \frac{-11.02}{50} = -0.2204 y=βˆ’60βˆ’48.9850=βˆ’108.9850=βˆ’2.1796y = \frac{-60 - 48.98}{50} = \frac{-108.98}{50} = -2.1796

Conclusion

In this article, we have solved the quadratic equation (5y+6)2=24(5y + 6)^2 = 24 using the quadratic formula. We have found two possible values of yy, which are y=βˆ’6+265y = \frac{-6 + 2\sqrt{6}}{5} and y=βˆ’6βˆ’265y = \frac{-6 - 2\sqrt{6}}{5}. These values are the solutions to the quadratic equation.

Final Answer

The final answer is:

A. y=βˆ’6+265y = \frac{-6 + 2\sqrt{6}}{5} and y=βˆ’6βˆ’265y = \frac{-6 - 2\sqrt{6}}{5}

Introduction

In our previous article, we solved the quadratic equation (5y+6)2=24(5y + 6)^2 = 24 using the quadratic formula. In this article, we will provide a Q&A section to help clarify any doubts or questions that readers may have.

Q: What is the quadratic equation?

A: The 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 ax2+bx+c=0ax^2 + bx + c = 0, where aa, bb, and cc are constants, and xx is the variable.

Q: How do I solve a quadratic equation?

A: To solve a quadratic equation, you can use the quadratic formula, which is given by:

x=βˆ’bΒ±b2βˆ’4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

You can also use other methods, such as factoring or completing the square, depending on the specific equation.

Q: What is the quadratic formula?

A: The quadratic formula is a mathematical formula that is used to solve quadratic equations. It is given by:

x=βˆ’bΒ±b2βˆ’4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Q: How do I use the quadratic formula?

A: To use the quadratic formula, you need to plug in the values of aa, bb, and cc into the formula. You also need to simplify the expression inside the square root.

Q: What is the expression inside the square root?

A: The expression inside the square root is given by b2βˆ’4acb^2 - 4ac. This expression is also known as the discriminant.

Q: What is the discriminant?

A: The discriminant is the expression inside the square root of the quadratic formula. It is given by b2βˆ’4acb^2 - 4ac. If the discriminant is positive, then the quadratic equation has two distinct solutions. If the discriminant is zero, then the quadratic equation has one repeated solution. If the discriminant is negative, then the quadratic equation has no real solutions.

Q: How do I simplify the quadratic formula?

A: To simplify the quadratic formula, you need to simplify the expression inside the square root. You can do this by factoring or using other mathematical techniques.

Q: What are the solutions to the quadratic equation (5y+6)2=24(5y + 6)^2 = 24?

A: The solutions to the quadratic equation (5y+6)2=24(5y + 6)^2 = 24 are given by:

y=βˆ’6+265y = \frac{-6 + 2\sqrt{6}}{5} and y=βˆ’6βˆ’265y = \frac{-6 - 2\sqrt{6}}{5}

Q: How do I check my solutions?

A: To check your solutions, you can plug them back into the original equation to see if they satisfy the equation.

Q: What are some common mistakes to avoid when solving quadratic equations?

A: Some common mistakes to avoid when solving quadratic equations include:

  • Not simplifying the expression inside the square root
  • Not using the correct values of aa, bb, and cc
  • Not checking the solutions
  • Not using the quadratic formula correctly

Conclusion

In this article, we have provided a Q&A section to help clarify any doubts or questions that readers may have about solving quadratic equations. We have also provided some common mistakes to avoid when solving quadratic equations.

Final Answer

The final answer is:

A. y=βˆ’6+265y = \frac{-6 + 2\sqrt{6}}{5} and y=βˆ’6βˆ’265y = \frac{-6 - 2\sqrt{6}}{5}