Solve The Quadratic Equation: $4x^2 + 17x + 15 = 0$

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Introduction

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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: $4x^2 + 17x + 15 = 0$. We will use various methods to find the solutions, including factoring, the quadratic formula, and completing the square.

What is a Quadratic Equation?

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A quadratic equation is a polynomial equation of degree two, which means the highest power of the variable (in this case, x) is two. The general form of a quadratic equation is:

$$ax^2 + bx + c = 0$$

where a, b, and c are constants, and a is not equal to zero.

The Quadratic Formula

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The quadratic formula is a powerful tool for solving quadratic equations. It is given by:

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

This formula will give us two solutions for the equation, which may be real or complex numbers.

Solving the Quadratic Equation: $4x^2 + 17x + 15 = 0$

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Factoring the Quadratic Equation

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To factor the quadratic equation, we need to find two numbers whose product is $4 \times 15 = 60$ and whose sum is 17. These numbers are 20 and 3, since $20 \times 3 = 60$ and $20 + 3 = 23$, which is not correct, but $-20 \times -3 = 60$ and $-20 + (-3) = -23$, which is also not correct, but $-20 \times 3 = -60$ and $-20 + 3 = -17$, which is correct.

So, we can write the quadratic equation as:

$$(2x + 3)(2x - 5) = 0$$

This tells us that either $2x + 3 = 0$ or $2x - 5 = 0$.

Solving the First Factor: $2x + 3 = 0$

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To solve the first factor, we can isolate x by subtracting 3 from both sides and then dividing by 2:

$$2x = -3$$

$$x = \frac{-3}{2}$$

Solving the Second Factor: $2x - 5 = 0$

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To solve the second factor, we can isolate x by adding 5 to both sides and then dividing by 2:

$$2x = 5$$

$$x = \frac{5}{2}$$

Using the Quadratic Formula

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We can also use the quadratic formula to solve the equation. Plugging in the values of a, b, and c, we get:

$$x = \frac{-17 \pm \sqrt{17^2 - 4 \times 4 \times 15}}{2 \times 4}$$

$$x = \frac{-17 \pm \sqrt{289 - 240}}{8}$$

$$x = \frac{-17 \pm \sqrt{49}}{8}$$

$$x = \frac{-17 \pm 7}{8}$$

This gives us two solutions:

$$x = \frac{-17 + 7}{8} = \frac{-10}{8} = \frac{-5}{4}$$

$$x = \frac{-17 - 7}{8} = \frac{-24}{8} = -3$$

Completing the Square

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We can also solve the equation by completing the square. To do this, we need to rewrite the equation in the form:

$$x^2 + bx + c = 0$$

We can do this by dividing the entire equation by 4:

$$x^2 + \frac{17}{4}x + \frac{15}{4} = 0$$

Next, we need to find a number that we can add to both sides of the equation to make the left-hand side a perfect square. This number is $\left(\frac{17}{8}\right)^2 = \frac{289}{64}$.

Adding this number to both sides, we get:

$$x^2 + \frac{17}{4}x + \frac{289}{64} = \frac{15}{4} + \frac{289}{64}$$

This can be rewritten as:

$$\left(x + \frac{17}{8}\right)^2 = \frac{15}{4} + \frac{289}{64}$$

Simplifying the right-hand side, we get:

$$\left(x + \frac{17}{8}\right)^2 = \frac{120 + 289}{64}$$

$$\left(x + \frac{17}{8}\right)^2 = \frac{409}{64}$$

Taking the square root of both sides, we get:

$$x + \frac{17}{8} = \pm \sqrt{\frac{409}{64}}$$

Simplifying the right-hand side, we get:

$$x + \frac{17}{8} = \pm \frac{\sqrt{409}}{8}$$

Subtracting $\frac{17}{8}$ from both sides, we get:

$$x = -\frac{17}{8} \pm \frac{\sqrt{409}}{8}$$

This gives us two solutions:

$$x = -\frac{17}{8} + \frac{\sqrt{409}}{8}$$

$$x = -\frac{17}{8} - \frac{\sqrt{409}}{8}$$

Conclusion

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In this article, we have solved the quadratic equation $4x^2 + 17x + 15 = 0$ using various methods, including factoring, the quadratic formula, and completing the square. We have found that the solutions are $x = \frac{-5}{4}$ and $x = -3$. These solutions can be verified by plugging them back into the original equation.

Final Answer

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The final answer is $\boxed{\frac{-5}{4}, -3}$.

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Introduction

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Quadratic equations are a fundamental concept in mathematics, and solving them can be a challenging task. In this article, we will answer some of the most frequently asked questions about quadratic equations, including how to solve them, what the quadratic formula is, and how to use it.

Q: What is a Quadratic Equation?

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A: A quadratic equation is a polynomial equation of degree two, which means the highest power of the variable (in this case, x) is two. The general form of a quadratic equation is:

$$ax^2 + bx + c = 0$$

where a, b, and c are constants, and a is not equal to zero.

Q: What is the Quadratic Formula?

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A: The quadratic formula is a powerful tool for solving quadratic equations. It is given by:

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

This formula will give us two solutions for the equation, which may be real or complex numbers.

Q: How Do I Use the Quadratic Formula?

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A: To use the quadratic formula, you need to plug in the values of a, b, and c from the quadratic equation into the formula. For example, if we have the equation $x^2 + 5x + 6 = 0$, we can plug in the values a = 1, b = 5, and c = 6 into the formula:

$$x = \frac{-5 \pm \sqrt{5^2 - 4 \times 1 \times 6}}{2 \times 1}$$

$$x = \frac{-5 \pm \sqrt{25 - 24}}{2}$$

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

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

This gives us two solutions:

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

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

Q: What is the Difference Between the Quadratic Formula and Factoring?

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A: The quadratic formula and factoring are two different methods for solving quadratic equations. The quadratic formula is a general method that can be used to solve any quadratic equation, while factoring is a specific method that can be used to solve quadratic equations that can be factored.

For example, if we have the equation $x^2 + 5x + 6 = 0$, we can factor it as:

$$(x + 3)(x + 2) = 0$$

This tells us that either $x + 3 = 0$ or $x + 2 = 0$.

Q: When Should I Use the Quadratic Formula and When Should I Use Factoring?

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A: You should use the quadratic formula when the quadratic equation cannot be factored, or when you are not sure if it can be factored. You should use factoring when the quadratic equation can be factored, or when you want to check your answer.

Q: Can I Use the Quadratic Formula to Solve Quadratic Equations with Complex Solutions?

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A: Yes, you can use the quadratic formula to solve quadratic equations with complex solutions. The quadratic formula will give you two solutions, which may be real or complex numbers.

For example, if we have the equation $x^2 + 1 = 0$, we can plug in the values a = 1, b = 0, and c = 1 into the formula:

$$x = \frac{-0 \pm \sqrt{0^2 - 4 \times 1 \times 1}}{2 \times 1}$$

$$x = \frac{0 \pm \sqrt{0 - 4}}{2}$$

$$x = \frac{0 \pm \sqrt{-4}}{2}$$

$$x = \frac{0 \pm 2i}{2}$$

$$x = \pm i$$

This gives us two complex solutions:

$$x = i$$

$$x = -i$$

Conclusion

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In this article, we have answered some of the most frequently asked questions about quadratic equations, including how to solve them, what the quadratic formula is, and how to use it. We have also discussed the difference between the quadratic formula and factoring, and when to use each method.

Final Answer

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The final answer is $\boxed{yes}$, you can use the quadratic formula to solve quadratic equations with complex solutions.