Quadratic equations are a fundamental part of algebra and often arise in various math problems. In this article, we will explore how to solve the quadratic equation x² – 11x + 28 = 0 – 11x + 28 = 0x2−11x+28=0 through factorization. We will also provide a step-by-step guide and explanation, including a video link for additional clarity.

This quadratic equation follows a common pattern, and once you understand the process of factorization, solving similar problems becomes much easier. Let’s break it down!

## Understanding the Quadratic Equation

A quadratic equation is any equation of the form:ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0

Where:

- aaa, bbb, and ccc are constants,
- xxx represents an unknown variable.

In the given equation x2−11x+28=0x^2 – 11x + 28 = 0x2−11x+28=0, the coefficients are as follows:

- a=1a = 1a=1,
- b=−11b = -11b=−11,
- c=28c = 28c=28.

Quadratic equations typically have two solutions, and these solutions are the points where the equation crosses the x-axis (if plotted on a graph). The factorization method is a common technique to solve these types of equations.

## How to Factorize x2−11x+28=0x^2 – 11x + 28 = 0x2−11x+28=0

Now that we understand the form of the equation, let’s move on to solving it by factorization. Factorization involves breaking down the equation into two simpler binomials that, when multiplied, give the original quadratic equation.

### Step-by-Step Guide:

**Identify the coefficients**:- In x2−11x+28=0x^2 – 11x + 28 = 0x2−11x+28=0, we have:
- a=1a = 1a=1,
- b=−11b = -11b=−11,
- c=28c = 28c=28.

- In x2−11x+28=0x^2 – 11x + 28 = 0x2−11x+28=0, we have:
**Find two numbers that multiply to give ccc (28) and add to give bbb (-11)**:- We are looking for two numbers that multiply to 282828 and add up to −11-11−11.
- The numbers −7-7−7 and −4-4−4 satisfy these conditions because: −7×−4=28and−7+(−4)=−11-7 \times -4 = 28 \quad \text{and} \quad -7 + (-4) = -11−7×−4=28and−7+(−4)=−11

**Rewrite the middle term using these numbers**:- We can now rewrite the equation by splitting the middle term (-11x): x2−7x−4x+28=0x^2 – 7x – 4x + 28 = 0x2−7x−4x+28=0

**Group the terms**:- Next, group the terms into pairs: (x2−7x)−(4x−28)=0(x^2 – 7x) – (4x – 28) = 0(x2−7x)−(4x−28)=0

**Factor out the common terms**:- Factor each group: x(x−7)−4(x−7)=0x(x – 7) – 4(x – 7) = 0x(x−7)−4(x−7)=0

**Combine the common factors**:- Since x−7x – 7x−7 is common in both terms, we can factor it out: (x−7)(x−4)=0(x – 7)(x – 4) = 0(x−7)(x−4)=0

**Solve for xxx**:- Now that we have the factored form, we can find the values of xxx by setting each factor equal to zero: x−7=0orx−4=0x – 7 = 0 \quad \text{or} \quad x – 4 = 0x−7=0orx−4=0
- Therefore, the solutions are: x=7andx=4x = 7 \quad \text{and} \quad x = 4x=7andx=4

## Verifying the Factorization

After factorizing, it’s always a good idea to check if the solutions are correct. To verify:

- Substitute x=7x = 7x=7 and x=4x = 4x=4 into the original equation x2−11x+28=0x^2 – 11x + 28 = 0x2−11x+28=0.

For x=7x = 7x=7:72−11(7)+28=49−77+28=07^2 – 11(7) + 28 = 49 – 77 + 28 = 072−11(7)+28=49−77+28=0

So, x=7x = 7x=7 is a correct solution.

For x=4x = 4x=4:42−11(4)+28=16−44+28=04^2 – 11(4) + 28 = 16 – 44 + 28 = 042−11(4)+28=16−44+28=0

So, x=4x = 4x=4 is also a correct solution.

## Related Video Explanation

For those who prefer a visual guide, here is a recommended video that explains how to solve and factorize quadratic equations, specifically addressing the equation x2−11x+28=0x^2 – 11x + 28 = 0x2−11x+28=0:

This video walks you through the factorization process, offering additional insights and examples to enhance your understanding.

### Conclusion

In conclusion, the quadratic equation x2−11x+28=0x^2 – 11x + 28 = 0x2−11x+28=0 can be easily solved using factorization. By identifying two numbers that multiply to the constant term and add up to the coefficient of the middle term, we can break down the equation into two binomials. In this case, the solutions are x=7x = 7x=7 and x=4x = 4x=4.

Learning to factorize quadratic equations like this one is an essential skill in algebra, and with practice, it becomes an intuitive and effective method for solving similar problems.

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