February 21, 2024

Solve – x2-11x+28=0 | Methods and Solutions

In this article, we’ll delve into solving the specific quadratic equation and explore its implications. Quadratic equations are a fundamental part of algebra, representing a parabolic relationship in a variety of real-world scenarios. Let’s start with the solutions.

The Quadratic Equation –

The general form of a quadratic equation is . In our case, the equation can be identified with , , and .

Method 1: Factoring

The equation can be factored if we can find two numbers that multiply to 28 (the constant term) and add up to -11 (the coefficient of x). These numbers are -7 and -4. Therefore, the equation can be factored as:

Setting each factor equal to zero gives us the solutions:

Hence, the solutions are and .

Method 2: Completing the Square

To solve by completing the square, rearrange the equation and complete the square on the left side:

  1. Rearrange the equation:
  2. Add the square of half the coefficient of x to both sides: x^211x +(112)2=28+(112)2  x^211x+30.25=2.25
  3. Write the left side as a perfect square: (x5.5)2=2.25
  4. Take the square root of both sides:  x 5.5=±2.25
  5. Solve for x: 

Thus, the solutions are and .

Method 3: Quadratic Formula

The quadratic formula is =−b ±b24ac2a, where , , and are coefficients from the quadratic equation ax^2+bx+c=0. For our equation:

Plugging these values into the formula:





Therefore, the solutions are:



The solutions for are and , irrespective of the method used

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Graphical Interpretation

The graph of is a parabola opening upwards. The roots of the equation, and , represent the points where this parabola intersects the x-axis.


Solving quadratic equations like can be approached through various methods, each providing a unique perspective on the problem. The roots of this particular equation, 4 and 7, offer insights into the parabolic nature of quadratic relationships in mathematics and their applications.

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