Tenth Grade Mathematics: Solving by Completing the Square

While splitting the middle term is fast when rational factors are readily visible, it fails when the roots involve non-trivial radicals or fractions. In contrast, the Method of Completing the Square is a universal geometric and algebraic technique that solves any quadratic equation regardless of its roots. Originally conceived geometrically by ancient Babylonian and Islamic mathematicians like Al-Khwarizmi, this method transforms an arbitrary quadratic expression into a perfect square binomial, allowing solutions to be extracted directly by taking square roots.

The Geometric Foundation of Completing the Square

The name "completing the square" reflects a physical geometric construction. Consider the expression x^2 + 2bx:

Geometry of x^2 + 2*b*x:
We have a square of side x (area x^2) and two rectangles of sides x and b (area 2*b*x):

     +------------+------+
     |            |      |
     |   x * x    | x*b  |
   x |   = x^2    |      |
     |            |      |
     +------------+------+
     |   b * x    | ???? | <--- MISSING CORNER PIECE OF AREA b^2!
   b |            |      |
     +------------+------+
           x         b

To complete the large square of side (x + b), we must add the missing corner
square of area b^2!
Total completed area: (x + b)^2 = x^2 + 2*b*x + b^2

Algebraically, whenever we possess x^2 + k*x, we complete the square by adding the square of half the linear coefficient: (k/2)^2.

The Algorithmic Step-by-Step Procedure

To solve any quadratic equation ax^2 + bx + c = 0 by completing the square:

Step 1: Divide the entire equation by the leading coefficient a so that the
        coefficient of x^2 becomes 1:
        x^2 + (b/a)*x + (c/a) = 0

Step 2: Transpose the constant term to the right-hand side:
        x^2 + (b/a)*x = -c/a

Step 3: Identify the coefficient of x, which is (b/a).
        Calculate half of this coefficient: (1/2) * (b/a) = b / (2*a).
        Calculate its square: [b / (2*a)]^2 = b^2 / (4*a^2).

Step 4: Add this square to BOTH sides of the equation:
        x^2 + (b/a)*x + [b / (2*a)]^2 = -c/a + [b / (2*a)]^2

Step 5: Write the left-hand side as a perfect square binomial:
        [x + b / (2*a)]^2 = (b^2 - 4*a*c) / (4*a^2)

Step 6: Take the square root of both sides (incorporating +/-) and solve for x:
        x + b / (2*a) = +/- sqrt(b^2 - 4*a*c) / (2*a)
        x = -b / (2*a) +/- sqrt(b^2 - 4*a*c) / (2*a)

Notice that executing this procedure on the general equation ax^2 + bx + c = 0 produces the universal Quadratic Formula directly!

Detailed Worked Example

Solve the quadratic equation by the method of completing the square:

2*x^2 - 5*x + 3 = 0

Step 1: Divide by leading coefficient 2:

x^2 - (5/2)*x + 3/2 = 0

Step 2: Transpose constant term to right side:

x^2 - (5/2)*x = -3/2

Step 3: Calculate half of the linear coefficient (-5/2):

Half = (1/2) * (-5/2) = -5/4
Square = (-5/4)^2 = 25/16

Step 4: Add 25/16 to both sides:

x^2 - (5/2)*x + 25/16 = -3/2 + 25/16

Step 5: Express left side as a perfect square and simplify right side:

(x - 5/4)^2 = (-24 + 25) / 16
(x - 5/4)^2 = 1 / 16

Step 6: Take square roots of both sides:

x - 5/4 = +/- sqrt(1 / 16)
x - 5/4 = +/- 1/4

Case 1: x = 5/4 + 1/4 = 6/4 = 3/2
Case 2: x = 5/4 - 1/4 = 4/4 = 1

The roots of the equation are x = 3/2 and x = 1.

Chapter Exercises

Exercise 1: Solve the quadratic equation by completing the square: x^2 + 4*x - 5 = 0.

Exercise 2: Solve by completing the square: 5x^2 - 6x - 2 = 0. Show all intermediate fractional steps.

Exercise 3: Show by completing the square that the equation 2x^2 - 4x + 3 = 0 has no real roots.

Solutions and Explanations

Solution 1: We have x^2 + 4x - 5 = 0. Transpose constant term: x^2 + 4x = 5. The coefficient of x is 4; half of 4 is 2; square of 2 is 4. Add 4 to both sides: x^2 + 4*x + 4 = 5 + 4 ==> (x + 2)^2 = 9. Take square root of both sides: x + 2 = +/- sqrt(9) = +/- 3. Case 1: x + 2 = 3 ==> x = 1. Case 2: x + 2 = -3 ==> x = -5. The roots are x = 1 and x = -5.

Solution 2: For 5x^2 - 6x - 2 = 0, divide by 5: x^2 - (6/5)x - 2/5 = 0. Transpose constant: x^2 - (6/5)x = 2/5. Coefficient of x is -6/5; half of -6/5 is -3/5; square is 9/25. Add 9/25 to both sides: x^2 - (6/5)*x + 9/25 = 2/5 + 9/25 ==> (x - 3/5)^2 = (10 + 9)/25 = 19/25. Take square roots: x - 3/5 = +/- sqrt(19)/5. Isolate x: x = (3 +/- sqrt(19))/5. The two real roots are x = (3 + sqrt(19))/5 and x = (3 - sqrt(19))/5.

Solution 3: Given 2x^2 - 4x + 3 = 0. Divide by 2: x^2 - 2x + 3/2 = 0. Transpose constant: x^2 - 2x = -3/2. Half of -2 is -1; square of -1 is 1. Add 1 to both sides: x^2 - 2*x + 1 = -3/2 + 1 ==> (x - 1)^2 = -1/2. Because the left-hand side is the square of a real number, (x - 1)^2 >= 0 for all real values of x. However, the right-hand side is -1/2, which is strictly negative. Since the square of a real number can never be negative, no real value of x satisfies this equality. Therefore, the equation has no real roots.