Solve $5d^2 = 4(4d - 3$\].

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Introduction

In this article, we will delve into solving a quadratic equation of the form $5d^2 = 4(4d - 3)$. This equation is a quadratic equation, which is a polynomial equation of degree two, meaning the highest power of the variable is two. Quadratic equations are commonly encountered in various fields, including algebra, geometry, and physics. The solution to this equation will provide us with the values of the variable $d$ that satisfy the given equation.

Understanding the Equation

The given equation is $5d^2 = 4(4d - 3)$. To solve this equation, we need to isolate the variable $d$ on one side of the equation. The first step is to expand the right-hand side of the equation using the distributive property. This will give us $5d^2 = 16d - 12$.

Expanding the Right-Hand Side

To expand the right-hand side of the equation, we need to multiply the constant term $4$ by the expression inside the parentheses $4d - 3$. This will give us $16d - 12$.

Simplifying the Equation

Now that we have expanded the right-hand side of the equation, we can simplify the equation by combining like terms. The equation becomes $5d^2 - 16d + 12 = 0$.

Solving the Quadratic Equation

To solve the quadratic equation $5d^2 - 16d + 12 = 0$, we can use the quadratic formula. The quadratic formula is given by $d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a$, $b$, and $c$ are the coefficients of the quadratic equation.

Applying the Quadratic Formula

In our equation, $a = 5$, $b = -16$, and $c = 12$. Plugging these values into the quadratic formula, we get $d = \frac{-(-16) \pm \sqrt{(-16)^2 - 4(5)(12)}}{2(5)}$.

Simplifying the Quadratic Formula

Simplifying the expression inside the square root, we get $d = \frac{16 \pm \sqrt{256 - 240}}{10}$.

Evaluating the Square Root

Evaluating the square root, we get $d = \frac{16 \pm \sqrt{16}}{10}$.

Simplifying the Expression

Simplifying the expression, we get $d = \frac{16 \pm 4}{10}$.

Finding the Solutions

Now that we have simplified the expression, we can find the solutions to the equation. The two possible solutions are $d = \frac{16 + 4}{10}$ and $d = \frac{16 - 4}{10}$.

Calculating the Solutions

Calculating the solutions, we get $d = \frac{20}{10}$ and $d = \frac{12}{10}$.

Simplifying the Solutions

Simplifying the solutions, we get $d = 2$ and $d = \frac{6}{5}$.

Conclusion

In this article, we have solved the quadratic equation $5d^2 = 4(4d - 3)$. We first expanded the right-hand side of the equation, then simplified the equation by combining like terms. We then applied the quadratic formula to find the solutions to the equation. The two possible solutions are $d = 2$ and $d = \frac{6}{5}$. These solutions satisfy the given equation and provide us with the values of the variable $d$ that satisfy the equation.

Final Answer

The final answer is $\boxed{2, \frac{6}{5}}$.

Related Topics

• Quadratic equations
• Algebra
• Geometry
• Physics

References

• [1] "Quadratic Equations" by Math Open Reference
• [2] "Algebra" by Khan Academy
• [3] "Geometry" by Math Is Fun
• [4] "Physics" by Physics Classroom
  

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Introduction

In our previous article, we solved the quadratic equation $5d^2 = 4(4d - 3)$. In this article, we will provide a Q&A section to help clarify any doubts or questions that readers may have. We will cover various topics related to the solution of the quadratic equation, including the quadratic formula, simplifying expressions, and finding solutions.

Q: What is the quadratic formula?

A: The quadratic formula is a mathematical formula used to find the solutions to a quadratic equation of the form $ax^2 + bx + c = 0$. The formula is given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a$, $b$, and $c$ are the coefficients of the quadratic equation.

Q: How do I apply the quadratic formula to solve a quadratic equation?

A: To apply the quadratic formula, you need to identify the coefficients $a$, $b$, and $c$ in the quadratic equation. Then, plug these values into the quadratic formula and simplify the expression to find the solutions.

Q: What is the difference between the quadratic formula and factoring?

A: The quadratic formula and factoring are two different methods used to solve quadratic equations. Factoring involves expressing the quadratic equation as a product of two binomials, while the quadratic formula involves using a formula to find the solutions.

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

A: Yes, the quadratic formula can be used to solve any quadratic equation of the form $ax^2 + bx + c = 0$. However, the formula may not always yield real solutions, especially if the expression inside the square root is negative.

Q: How do I simplify expressions involving square roots?

A: To simplify expressions involving square roots, you need to identify the square root and then simplify the expression inside the square root. You can also use the properties of square roots, such as the fact that $\sqrt{a^2} = a$, to simplify the expression.

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. These values can be used to solve problems in various fields, including algebra, geometry, and physics.

Q: Can I use the quadratic formula to solve quadratic equations with complex coefficients?

A: Yes, the quadratic formula can be used to solve quadratic equations with complex coefficients. However, the solutions may involve complex numbers, which can be represented in the form $a + bi$, where $a$ and $b$ are real numbers and $i$ is the imaginary unit.

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

A: To determine the number of solutions to a quadratic equation, you need to examine the discriminant, which is the expression inside 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: Can I use the quadratic formula to solve quadratic equations with rational coefficients?

A: Yes, the quadratic formula can be used to solve quadratic equations with rational coefficients. However, the solutions may involve rational numbers, which can be represented in the form $\frac{a}{b}$, where $a$ and $b$ are integers.

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 using the quadratic formula. We have covered various topics related to the solution of quadratic equations, including the quadratic formula, simplifying expressions, and finding solutions.

Final Answer

The final answer is $\boxed{2, \frac{6}{5}}$.

Related Topics

• Quadratic equations
• Algebra
• Geometry
• Physics

References

• [1] "Quadratic Equations" by Math Open Reference
• [2] "Algebra" by Khan Academy
• [3] "Geometry" by Math Is Fun
• [4] "Physics" by Physics Classroom