Solving Quadratic Systems: A Step-by-Step Guide

by Alex Johnson 48 views

In the realm of algebra, solving systems of equations is a fundamental skill. When these systems involve quadratic equations, the process requires careful attention and a systematic approach. This guide provides a detailed walkthrough on how to solve a system where one equation is quadratic and the other is linear, using the specific example:

  • y = x^2 + 17x + 86
  • y = -2x - 4

Step 1: Setting the Equations Equal

The cornerstone of solving this system lies in recognizing that both equations are expressed in terms of 'y'. This allows us to equate the two expressions, effectively eliminating 'y' and creating a single equation in terms of 'x'. By setting the quadratic expression equal to the linear expression, we pave the way for solving for the 'x' values that satisfy both equations simultaneously. This is a crucial first step that simplifies the problem into a manageable quadratic equation.

When we equate the two equations, we get:

  • x^2 + 17x + 86 = -2x - 4

This equation now represents the points where the parabola and the line intersect. Solving for 'x' will give us the x-coordinates of these intersection points.

Step 2: Rearranging to Form a Quadratic Equation

To solve for 'x', we need to rearrange the equation into the standard quadratic form: ax^2 + bx + c = 0. This involves moving all terms to one side of the equation, leaving zero on the other side. By doing so, we prepare the equation for factoring, completing the square, or using the quadratic formula – all standard techniques for solving quadratic equations.

Adding 2x and 4 to both sides of the equation, we obtain:

  • x^2 + 17x + 2x + 86 + 4 = 0
  • x^2 + 19x + 90 = 0

Now, we have a standard quadratic equation that we can solve using various methods.

Step 3: Solving the Quadratic Equation

With the quadratic equation in standard form (x^2 + 19x + 90 = 0), we have several options for solving it. These include factoring, completing the square, or using the quadratic formula. In this case, factoring is the most straightforward approach. We look for two numbers that multiply to 90 and add up to 19. These numbers are 9 and 10.

Therefore, we can factor the quadratic equation as follows:

  • (x + 9)(x + 10) = 0

Setting each factor equal to zero gives us the solutions for 'x':

  • x + 9 = 0 => x = -9
  • x + 10 = 0 => x = -10

These are the x-coordinates of the points where the parabola and the line intersect.

Step 4: Finding the Corresponding 'y' Values

Now that we have the 'x' values, we need to find the corresponding 'y' values. We can do this by substituting each 'x' value back into either of the original equations. The linear equation (y = -2x - 4) is generally easier to work with. Let's substitute each 'x' value into this equation:

For x = -9:

  • y = -2(-9) - 4
  • y = 18 - 4
  • y = 14

For x = -10:

  • y = -2(-10) - 4
  • y = 20 - 4
  • y = 16

Thus, we have two solutions: (-9, 14) and (-10, 16).

Step 5: Verifying the Solutions

To ensure the accuracy of our solutions, it's crucial to verify them by substituting both the 'x' and 'y' values into both original equations. This step confirms that the solutions satisfy both equations simultaneously, ensuring that they are indeed the correct intersection points.

For the point (-9, 14):

  • Equation 1: 14 = (-9)^2 + 17(-9) + 86 => 14 = 81 - 153 + 86 => 14 = 14 (True)
  • Equation 2: 14 = -2(-9) - 4 => 14 = 18 - 4 => 14 = 14 (True)

For the point (-10, 16):

  • Equation 1: 16 = (-10)^2 + 17(-10) + 86 => 16 = 100 - 170 + 86 => 16 = 16 (True)
  • Equation 2: 16 = -2(-10) - 4 => 16 = 20 - 4 => 16 = 16 (True)

Both solutions satisfy both equations, confirming that they are correct.

Conclusion

Solving a system of equations involving a quadratic and a linear equation requires a systematic approach. By setting the equations equal, rearranging to form a standard quadratic equation, solving for 'x', finding the corresponding 'y' values, and verifying the solutions, we can accurately determine the points of intersection. This method provides a robust framework for solving similar problems in algebra.

In summary, the solutions to the system of equations are (-9, 14) and (-10, 16). This comprehensive guide provides a clear and concise method for solving such systems, reinforcing the fundamental principles of algebraic manipulation and problem-solving.

For further information on solving systems of equations, you can visit Khan Academy's resource on systems of equations.