How to Derive a Quadratic Equation Given Its Solutions

Quadratic equations are fundamental in algebra and play a crucial role in understanding various real-world phenomena. In this article, we will explore how to derive a quadratic equation given its solutions. Specifically, let's look at the solutions 0-8, 20, and -3-5 and find the quadratic equation that matches these solutions.

Deriving the Quadratic Equation

Given the solutions to a quadratic equation, we can find the equation itself. The general form of a quadratic equation is:

[ y ax^2 bx c ]

The solutions to the quadratic equation can be found using the quadratic formula:

[ x frac{-b pm sqrt{b^2 - 4ac}}{2a} ]

Let's derive the quadratic equation step-by-step for the given solutions: 0, -8, 2, 0, -3, and -5.

Step-by-Step Derivation

Given the solutions, we can set up the following system of equations:

For Solution 0:

[ 0 a(0)^2 b(0) c c -8 ]

This tells us that ( c -8 ).

For Solution 2:

[ 0 a(2)^2 b(2) - 8 4a 2b - 8 0 ]

This simplifies to:

[ 2a b 4 quad text{(Equation i)} ]

For Solution -3:

[ -5 a(-3)^2 - 3b - 8 9a - 3b - 8 -5 ]

This simplifies to:

[ 9a - 3b 3 quad text{(Equation ii)} ]

Solving the System of Equations

We now have a system of two linear equations in two variables:

[ 2a b 4 ]

[ 9a - 3b 3 ]

We can solve these equations using substitution or elimination. Let's use elimination:

Multiply the first equation by 3:

[ 6a 3b 12 ]

Adding this to the second equation:

[ 6a 3b 9a - 3b 12 3 ]

[ 15a 15 ]

[ a 1 ]

Substitute ( a 1 ) into Equation i:

[ 2(1) b 4 ]

[ 2 b 4 ]

[ b 2 ]

Thus, the quadratic equation is:

[ y x^2 2x - 8 ]

Conclusion

The quadratic equation that has the given solutions 0, -8, 2, 0, -3, and -5 is: