Completing the Square

Author(s): 
Barnabas Hughes

A major goal for secondary school students in their study of elementary algebra is to understand, solve, and apply the quadratic equation.  Modern texts commonly begin with an application of the quadratic equation focused on the parabola.  This is a grand improvement over the text used by the author that began with the bald statement, “Quadratic Equations.  An equation of the second degree is called a quadratic equation.”  Immediately the student was plunged into the zero law (\( ab = 0 ,\) etc.), and how to solve quadratic equations by factoring.  Only after struggling through 73 exercises would the students be challenged with some practical applications.  Useful as is factoring, it is not the original way of solving quadratic equations.

The quadratic equation, as we know it today, was first discussed and taught by Muhammed ibn Musa al-Khwarizmi (fl. 815-850).  At the command of his Caliph, he collected all the material he could find on algebra and wrote the first text on the subject.  The title of his book contains the word algebra.  After brief attention to first degree equations and simple quadratics that required only square roots for their solution, he turned to quadratic equations.  While he did not use the word equation, the quadratic equation is correctly named:  it focuses on the dimensions of a square.  In fact, quadratic means square.  Al-Khwarizmi, as Muhammed is more commonly called, solved quadratic equations by the method we call today, completing the square.  Again, this phrase describes exactly what he did, as seen in the solution of his example that is the classical quadratic equation, \[ x^2 + 10x = 39 .\]

In the upper left section of the illustration below, the terms of the equation, \( x^2 \) and \( 10x ,\) are represented by geometric figures, a square and a rectangle.  The sum of their areas is given to be \( 39 .\)  In the lower left section of the illustration, the rectangle is cut into two parts that are attached to adjacent sides of the square.  The L-shaped result is then filled in with a smaller square that fills out or completes the larger square.  The addition increases the given size of the figure from \( 39\) to \( 39 + 25 = 64 .\)  The solution follows easily.

 

 

Exercises:  (1) \( x^2 + 12x = 108 ;\)  (2) \( x^2 + 6x = 55 ;\)  (3) \( x^2 + 8x = 65 ;\)  (4) \( x^2 + 4x = 60 .\)


Completing the Square - Quadratics Using Addition

Author(s): 
Barnabas Hughes

A major goal for secondary school students in their study of elementary algebra is to understand, solve, and apply the quadratic equation.  Modern texts commonly begin with an application of the quadratic equation focused on the parabola.  This is a grand improvement over the text used by the author that began with the bald statement, “Quadratic Equations.  An equation of the second degree is called a quadratic equation.”  Immediately the student was plunged into the zero law (\( ab = 0 ,\) etc.), and how to solve quadratic equations by factoring.  Only after struggling through 73 exercises would the students be challenged with some practical applications.  Useful as is factoring, it is not the original way of solving quadratic equations.

The quadratic equation, as we know it today, was first discussed and taught by Muhammed ibn Musa al-Khwarizmi (fl. 815-850).  At the command of his Caliph, he collected all the material he could find on algebra and wrote the first text on the subject.  The title of his book contains the word algebra.  After brief attention to first degree equations and simple quadratics that required only square roots for their solution, he turned to quadratic equations.  While he did not use the word equation, the quadratic equation is correctly named:  it focuses on the dimensions of a square.  In fact, quadratic means square.  Al-Khwarizmi, as Muhammed is more commonly called, solved quadratic equations by the method we call today, completing the square.  Again, this phrase describes exactly what he did, as seen in the solution of his example that is the classical quadratic equation, \[ x^2 + 10x = 39 .\]

In the upper left section of the illustration below, the terms of the equation, \( x^2 \) and \( 10x ,\) are represented by geometric figures, a square and a rectangle.  The sum of their areas is given to be \( 39 .\)  In the lower left section of the illustration, the rectangle is cut into two parts that are attached to adjacent sides of the square.  The L-shaped result is then filled in with a smaller square that fills out or completes the larger square.  The addition increases the given size of the figure from \( 39\) to \( 39 + 25 = 64 .\)  The solution follows easily.

 

 

Exercises:  (1) \( x^2 + 12x = 108 ;\)  (2) \( x^2 + 6x = 55 ;\)  (3) \( x^2 + 8x = 65 ;\)  (4) \( x^2 + 4x = 60 .\)


Completing the Square - Quadratics Using Subtraction

Author(s): 
Barnabas Hughes

Another form of the quadratic equation has a negative second term; the area of a rectangle is subtracted from the area of a square.  The geometric solution is a little more complex in that the two parts of the rectangle are taken from the square.  The first removal requires the addition of a small square before the second half can be removed, as shown below.  Consequently, the original geometric square is finally reduced in size to a smaller square. 

For the quadratic equation \( x^2 - 6x = 16 :\)  In the upper left section of the illustration below, the terms of the equation, \( x^2 \) and \( 6x ,\) are represented by geometric figures, a square and a rectangle.  The difference of their areas is given to be \( 16 .\)  The rectangle is divided into two parts, each of area \( 3x .\)  In the upper right section of the illustration, one of these parts is removed from the square of area \(x^2 ,\) leaving a rectangle of area \( (x-3)x .\)  The goal now is to remove from this rectangle the remaining rectangle of area \( 3x.\)  In the lower left section of the illustration, a \( 3\times 3\) square is attached to the rectangle, resulting in an L-shaped region that is \( 9\) units larger than the rectangle.  Now a rectangle of area \( 3x\) may be removed from the L-shaped region, leaving, on the one hand, a square of area \( (x-3)(x-3) ,\) and, on the other hand, a region of area \( 16 + 9 = 25 .\)  The solution to \( x^2 - 6x = 16 \) now follows easily.

 

 

Exercises:  (1)  \( x^2 - 10x = 24 ;\)   (2)  \( x^2 - 8x = 20 ;\)   (3)  \( x^2 - 4x = 12 .\)

 

In his instructions for solving quadratic equations, al-Khwarizmi wrote that if the coefficient of the squared term is not unity, then make it so by using the multiplicative inverse.  (These were not his words, just his thought and instructions in modern language.)  Hence, he provided for the case \( ax^2 + bx = c .\)  Finally, he never envisioned an equation set equal to zero or to a negative number.  An equation had to equal something positive.  Consequently, the geometric method will not solve \( x^2 + 8x + 65 = 0 .\)  This requires the algebraic method that was derived from al-Khwarizmi’s geometric method.