Solving Quadratic Equations with T-SQL

July 13, 2023

Solving Quadratic Equations with T-SQL Quadratic-Equations-800x587 I often find myself teaching mathematics to my daughter, and recently, she has been diving into the realm of Quadratic Equations. As we delved deeper into the subject, we realized that solving quadratic equations by hand and verifying the solutions can be quite cumbersome and time-consuming. Determined to streamline this process, I embarked on a journey to leverage the power of T-SQL for solving quadratic equations. Today, I am excited to share my learnings with all of you. In this blog post, we will explore a practical approach to solving quadratic equations using T-SQL. Let’s dive in!

Understanding Quadratic Equations

A quadratic equation is typically represented as ax^2 + bx + c = 0, where ‘a’, ‘b’, and ‘c’ are constants, and ‘x’ represents the unknown variable. The goal is to find the values of ‘x’ that satisfy the equation. To solve the quadratic equation, we can use the quadratic formula.

The quadratic formula states that the solutions to the quadratic equation ax^2 + bx + c = 0 can be calculated using the following formula: x = (-b ± √(b^2 – 4ac)) / (2a)

To solve quadratic equations using T-SQL, we can translate the quadratic formula directly into a SQL query. For example, if I have this Equation: x^2 – 5x + 6 = 0:

I can write T-SQL as follows:

DECLARE @a FLOAT, @b FLOAT, @c FLOAT;
SET @a = 2;
SET @b = -5;
SET @c = 3;
SELECT
x1 = (-@b + SQRT(POWER(@b, 2) - (4 * @a * @c))) / (2 * @a),
x2 = (-@b - SQRT(POWER(@b, 2) - (4 * @a * @c))) / (2 * @a);

The above will give a solution as x1 = 1.5 and x2 = 1.

While working with my daughter, we realize that the above equation is not enough as it does not take care of the few situations when the equation is not quadratic or has invalid coefficients. So eventually, we evolved to the following equation.

DECLARE @a FLOAT, @b FLOAT, @c FLOAT;
SET @a = 2;
SET @b = -5;
SET @c = 3;
-- Calculate the discriminant
DECLARE @discriminant FLOAT;
SET @discriminant = POWER(@b, 2) - (4 * @a * @c);
SELECT
  x1_result = CASE
                WHEN @a = 0 OR @a IS NULL THEN 'Not a quadratic equation'
                WHEN @b IS NULL OR @c IS NULL THEN 'Invalid coefficients'
                WHEN @discriminant < 0 THEN 'No real solutions'
                ELSE 'Real solutions'
              END,
  x1 = CASE
         WHEN @a <> 0 AND @discriminant >= 0 THEN (-@b + SQRT(@discriminant)) / (2 * @a)
         ELSE NULL
       END,
  x2 = CASE
         WHEN @a <> 0 AND @discriminant >= 0 THEN (-@b - SQRT(@discriminant)) / (2 * @a)
         ELSE NULL
       END;

I noway claim that my T-SQL solution is 100% correct. I might have missed a few things here and there. Kindly correct me by providing more guidance.

You can always reach out to me on LinkedIn.

Reference: Pinal Dave (https://blog.sqlauthority.com)

Mathematical Function, SQL Function

SQL SERVER – Writing Function for The Rule of 72

Calculating Extended Internal Rate of Return (XIRR) in SQL Server with T-SQL

2 Comments. Leave new

yusep
August 3, 2023 12:20 pm
The given code appears to be a SQL script that calculates the discriminant and the solutions
(x1 and x2) for a quadratic equation with coefficients @a, @b, and @c. However, there are a
couple of issues with the code:
1. The condition in the first CASE statement is not necessary and can be simplified:
WHEN @a = 0 OR @a IS NULL THEN ‘Not a quadratic equation’
Since @a is defined as a FLOAT variable, it can never be NULL. Therefore, the condition @a
IS NULL is redundant, and we can safely remove it:
WHEN @a = 0 THEN ‘Not a quadratic equation’
2. The code does not handle the case when @a, @b, or @c is NULL. If any of these coefficients
is NULL, the calculation for the discriminant and the solutions will produce NULL results. To
handle this situation, we should add a separate condition to check for NULL coefficients:
WHEN @a IS NULL OR @b IS NULL OR @c IS NULL THEN ‘Invalid coefficients’
DECLARE @a FLOAT, @b FLOAT, @c FLOAT;
SET @a = 2;
SET @b = -5;
SET @c = 3;
— Calculate the discriminant
DECLARE @discriminant FLOAT;
SET @discriminant = POWER(@b, 2) – (4 * @a * @c);
SELECT
x1_result = CASE
WHEN @a = 0 THEN ‘Not a quadratic equation’
WHEN @a IS NULL OR @b IS NULL OR @c IS NULL THEN ‘Invalid coefficients’
WHEN @discriminant < 0 THEN 'No real solutions'
ELSE 'Real solutions'
END,
x1 = CASE
WHEN @a 0 AND @discriminant >= 0 THEN (-@b + SQRT(@discriminant)) / (2 * @a)
ELSE NULL
END,
x2 = CASE
WHEN @a 0 AND @discriminant >= 0 THEN (-@b – SQRT(@discriminant)) / (2 * @a)
ELSE NULL
END;
This revised code handles the case when coefficients @a, @b, or @c are NULL and provides appropriate messages accordingly. It calculates the discriminant and determines the nature of the solutions (real or non-real) for the given quadratic equation.
Reply
- Pinal Dave
  August 3, 2023 8:34 pm
  Thank you Yusep.
  Reply