Tenth Grade Mathematics: The Quadratic Formula (Sridharacharya's Method)

In historical mathematics, Indian mathematicians made pioneering contributions to the theory of second-degree equations. Notably, the ancient eighth-century mathematician Sridharacharya developed a direct, comprehensive method for obtaining the roots of any quadratic equation without needing fractional transformations at early steps. Today, this universal result is celebrated across the globe as the Quadratic Formula. It guarantees that any quadratic equation can be solved directly by evaluating its coefficients through a single master expression.

Historical Derivation: Sridharacharya's Method

Sridharacharya's brilliant insight was to eliminate fractions during the completion of the square by multiplying the entire equation by 4*a at the outset.

Theorem (The Quadratic Formula):
The roots of the standard quadratic equation a*x^2 + b*x + c = 0 (where a != 0)
are given by:
                     -b +/- sqrt(b^2 - 4*a*c)
                x = --------------------------
                               2*a

Complete Step-by-Step Proof

Start with the general equation in standard form:

a*x^2 + b*x + c = 0

Multiply the entire equation by 4*a:

4*a * (a*x^2 + b*x + c) = 0
4*a^2 * x^2 + 4*a*b*x + 4*a*c = 0

Notice that the first term is a perfect square: 4a^2 * x^2 = (2ax)^2. The middle term can be written as twice a product: 4abx = 2 * (2ax) * b. Transpose the constant 4ac to the right-hand side:

(2*a*x)^2 + 2*(2*a*x)*b = -4*a*c

To complete the square on the left side, we need to add b^2. Add b^2 to both sides of the equation:

(2*a*x)^2 + 2*(2*a*x)*b + b^2 = b^2 - 4*a*c

The left-hand side is now identically the expanded square of the binomial (2ax + b):

(2*a*x + b)^2 = b^2 - 4*a*c

Provided that b^2 - 4ac >= 0, we can take the square root of both sides:

2*a*x + b = +/- sqrt(b^2 - 4*a*c)

Subtract b from both sides:

2*a*x = -b +/- sqrt(b^2 - 4*a*c)

Divide by 2*a (which is valid since a != 0):

x = [-b +/- sqrt(b^2 - 4*a*c)] / (2*a)

This completes the formal proof.

Systematic Execution Protocol

To solve any quadratic equation using the formula:

Execution Steps:
1. Write equation in standard form: a*x^2 + b*x + c = 0.
2. Identify numerical values of a, b, and c with their exact signs.
3. Compute the discriminant: D = b^2 - 4*a*c.
   - If D < 0, STOP: the equation has no real roots.
   - If D >= 0, continue to Step 4.
4. Substitute a, b, and sqrt(D) into x = (-b +/- sqrt(D)) / (2*a).
5. Separate the two cases (+ and -) and reduce fractions to lowest terms.

Worked Formula Example:
Solve: 3*x^2 - 5*x + 2 = 0.

Step 1: a = 3, b = -5, c = 2.
Step 2: D = b^2 - 4*a*c = (-5)^2 - 4*(3)*(2) = 25 - 24 = 1.
        Since D > 0, two distinct real roots exist.
Step 3: x = [-(-5) +/- sqrt(1)] / (2 * 3)
        x = (5 +/- 1) / 6
Step 4: Case 1: x = (5 + 1) / 6 = 6 / 6 = 1
        Case 2: x = (5 - 1) / 6 = 4 / 6 = 2 / 3
Roots: x = 1 and x = 2/3.

Chapter Exercises

Exercise 1: Find the roots of the quadratic equation 2x^2 - 7x + 3 = 0 using the quadratic formula.

Exercise 2: Solve the quadratic equation 4x^2 + 4sqrt(3)*x + 3 = 0 using the quadratic formula.

Exercise 3: Solve for x: 1/(x + 4) - 1/(x - 7) = 11/30 (where x != -4, 7) using the quadratic formula.

Solutions and Explanations

Solution 1: For 2x^2 - 7x + 3 = 0, identify coefficients: a = 2, b = -7, c = 3. Compute discriminant D: D = b^2 - 4ac = (-7)^2 - 4(2)(3) = 49 - 24 = 25. Since D > 0, roots are real and distinct. Applying the quadratic formula: x = [-b +/- sqrt(D)] / (2*a) = [-(-7) +/- sqrt(25)] / (2 * 2) = (7 +/- 5) / 4. For the plus sign: x = (7 + 5)/4 = 12/4 = 3. For the minus sign: x = (7 - 5)/4 = 2/4 = 1/2. The roots are x = 3 and x = 1/2.

Solution 2: For 4x^2 + 4sqrt(3)x + 3 = 0, coefficients are a = 4, b = 4sqrt(3), c = 3. Compute D: D = (4sqrt(3))^2 - 4(4)(3) = 16(3) - 48 = 48 - 48 = 0. Since D = 0, the equation has two equal real roots. Applying formula: x = [-4sqrt(3) +/- sqrt(0)] / (2 * 4) = -4sqrt(3) / 8 = -sqrt(3) / 2. Thus, the roots are repeated: x = -sqrt(3)/2, -sqrt(3)/2.

Solution 3: Simplify the left side: [(x - 7) - (x + 4)] / [(x + 4)(x - 7)] = 11/30 ==> (x - 7 - x - 4) / (x^2 - 3x - 28) = 11/30 ==> -11 / (x^2 - 3x - 28) = 11/30. Divide both numerators by 11: -1 / (x^2 - 3x - 28) = 1 / 30. Cross-multiply: -(30) = x^2 - 3x - 28 ==> x^2 - 3x - 28 + 30 = 0 ==> x^2 - 3x + 2 = 0. Coefficients: a = 1, b = -3, c = 2. Compute D: D = (-3)^2 - 4(1)(2) = 9 - 8 = 1. Applying quadratic formula: x = [-(-3) +/- sqrt(1)] / (21) = (3 +/- 1) / 2. Roots: x = (3 + 1)/2 = 2, and x = (3 - 1)/2 = 1. Both roots satisfy x != -4, 7. Thus, x = 1, 2.