Find All Real Solutions Of The Equation By Factoring.$\[ 5x^2 - 29x + 6 = 0 \\]x = ?

Mar 1, 2025 by ADMIN 85 views

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Introduction

In this article, we will explore the process of finding real solutions to a quadratic equation by factoring. Factoring is a powerful technique used to solve quadratic equations, and it is essential to understand how to apply it correctly. We will use the given equation $5x^2 - 29x + 6 = 0$ as an example to demonstrate the steps involved in factoring and finding the real solutions.

Understanding Quadratic Equations

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. In our given equation, $a = 5$ , $b = -29$ , and $c = 6$ .

Factoring Quadratic Equations

Factoring a quadratic equation involves expressing it as a product of two binomials. The factored form of a quadratic equation is $(x - r_1)(x - r_2) = 0$ , where $r_1$ and $r_2$ are the roots of the equation. To factor a quadratic equation, we need to find two numbers whose product is $ac$ and whose sum is $b$ . In our given equation, we need to find two numbers whose product is $5 \times 6 = 30$ and whose sum is $-29$ .

Finding the Factors

To find the factors, we can start by listing the factors of 30 and then checking which pair of factors adds up to -29. The factors of 30 are: 1, 2, 3, 5, 6, 10, 15, and 30. We can try different pairs of factors to see which one adds up to -29.

Possible Factors

1 and 30: $1 + 30 = 31$ (not equal to -29)
2 and 15: $2 + 15 = 17$ (not equal to -29)
3 and 10: $3 + 10 = 13$ (not equal to -29)
5 and 6: $5 + 6 = 11$ (not equal to -29)
-1 and -30: $-1 + (-30) = -31$ (not equal to -29)
-2 and -15: $-2 + (-15) = -17$ (not equal to -29)
-3 and -10: $-3 + (-10) = -13$ (not equal to -29)
-5 and -6: $-5 + (-6) = -11$ (not equal to -29)

However, we can also try to factor the equation by grouping. We can rewrite the equation as $(5x^2 - 6x) - (23x) = 0$ . Then, we can factor out the common terms: $x(5x - 6) - 23(x) = 0$ . Now, we can factor out the common term $x$ : $x(5x - 6 - 23) = 0$ . This simplifies to $x(5x - 29) = 0$ .

Factoring the Equation

Now, we can see that the equation can be factored as $(5x - 29)(x - 6) = 0$ . This means that either $5x - 29 = 0$ or $x - 6 = 0$ .

Solving for x

To solve for $x$ , we can set each factor equal to zero and solve for $x$ . For the first factor, we have $5x - 29 = 0$ . Adding 29 to both sides gives $5x = 29$ . Dividing both sides by 5 gives $x = \frac{29}{5}$ .

For the second factor, we have $x - 6 = 0$ . Adding 6 to both sides gives $x = 6$ .

Conclusion

In this article, we have demonstrated how to find the real solutions of a quadratic equation by factoring. We used the given equation $5x^2 - 29x + 6 = 0$ as an example and showed how to factor it by grouping. We then solved for $x$ by setting each factor equal to zero and solving for $x$ . The real solutions of the equation are $x = \frac{29}{5}$ and $x = 6$ .

Final Answer

The final answer is: $\boxed{\frac{29}{5}, 6}$

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Introduction

In our previous article, we explored the process of finding real solutions to a quadratic equation by factoring. Factoring is a powerful technique used to solve quadratic equations, and it is essential to understand how to apply it correctly. In this article, we will address some frequently asked questions (FAQs) on factoring quadratic equations.

Q&A

Q: What is factoring in quadratic equations?

A: Factoring in quadratic equations involves expressing the equation as a product of two binomials. The factored form of a quadratic equation is $(x - r_1)(x - r_2) = 0$ , where $r_1$ and $r_2$ are the roots of the equation.

Q: How do I factor a quadratic equation?

A: To factor a quadratic equation, you need to find two numbers whose product is $ac$ and whose sum is $b$ . You can start by listing the factors of $ac$ and then checking which pair of factors adds up to $b$ .

Q: What if I have a quadratic equation that cannot be factored easily?

A: If a quadratic equation cannot be factored easily, you can try using other methods such as the quadratic formula or completing the square.

Q: Can I use factoring to solve quadratic equations with complex roots?

A: Yes, you can use factoring to solve quadratic equations with complex roots. However, you need to be careful when working with complex numbers.

Q: How do I determine if a quadratic equation has real or complex roots?

A: To determine if a quadratic equation has real or complex roots, you can use the discriminant, which is given by the formula $b^2 - 4ac$ . If the discriminant is positive, the equation has two real roots. If the discriminant is zero, the equation has one real root. If the discriminant is negative, the equation has two complex roots.

Q: Can I use factoring to solve quadratic equations with rational coefficients?

A: Yes, you can use factoring to solve quadratic equations with rational coefficients. However, you need to be careful when working with rational numbers.

Q: How do I factor a quadratic equation with a negative leading coefficient?

A: To factor a quadratic equation with a negative leading coefficient, you can start by factoring out the negative sign. Then, you can factor the remaining expression as usual.

Q: Can I use factoring to solve quadratic equations with a coefficient of 1?

A: Yes, you can use factoring to solve quadratic equations with a coefficient of 1. However, you need to be careful when working with equations of this form.

Examples

Example 1: Factoring a Quadratic Equation with Real Roots

Factor the quadratic equation $x^2 + 5x + 6 = 0$ .

Solution: To factor this equation, we need to find two numbers whose product is $6$ and whose sum is $5$ . The numbers are $2$ and $3$ , so we can write the equation as $(x + 2)(x + 3) = 0$ . This means that either $x + 2 = 0$ or $x + 3 = 0$ . Solving for $x$ , we get $x = -2$ or $x = -3$ .

Example 2: Factoring a Quadratic Equation with Complex Roots

Factor the quadratic equation $x^2 + 4x + 5 = 0$ .

Solution: To factor this equation, we need to find two numbers whose product is $5$ and whose sum is $4$ . The numbers are $1$ and $5$ , so we can write the equation as $(x + 1)(x + 5) = 0$ . This means that either $x + 1 = 0$ or $x + 5 = 0$ . Solving for $x$ , we get $x = -1$ or $x = -5$ . However, since the discriminant is negative, the roots are complex.

Conclusion

In this article, we have addressed some frequently asked questions (FAQs) on factoring quadratic equations. We have provided examples and explanations to help you understand the process of factoring and solving quadratic equations. We hope this article has been helpful in clarifying any doubts you may have had on this topic.

Final Answer

The final answer is: $\boxed{No specific answer, as this is a Q&A article.}$