1. Completing the Square: This is a method involves rewriting a part of a quadratic equation as a perfect square and then taking the square root of both sides to find the solution. To begin with, lets look at a few examples the fairly easy quadratic equations to solve. The first one is x2 = 121. It may be tempting to just say that the answer is just 11, but this would be incomplete. To start with I will use the factoring method from the Solving Equations post:
x2 = 121
x2 - 121 = 0
(x - 11)(x + 11) = 0
x = 11, or -11
Another way to write this is x = ±11, read “x equals plus or minus 11”. Checking is very easy because 112 = 121 and (-11) 2 = 121. As you can see, 11 was only part of the solution. This is because squaring has the nice property that both negative and positive values turn out positive when the value is squared. Thus, whenever you take the square root of both sides of an equation you have to include both the positive and negative values so you must always introduce the ± in your answer or write two separate equations. To solve this equation without factoring you would write it as follows:
x2 = 121
√(x2) = ±√(121)
x = ±11
Now lets do another one:
(2x + 7)2 = 25
2x + 7 = ±5
2x = -7 ± 5
x = (-7 ± 5)/2
x = (-7 + 5)/2 or x = (-7 - 5)/2
x = -1, -6
Check: x = -1
(2(-1) + 7)2 = 25?
52 = 25? YES
Check: x = -6
(2(-6) + 7)2 = 25?
(-5)2 = 25? YES
So that is not so bad as long as the quadratic is written as a perfect square. But now we need to figure out how to solve an equation like 4x2 + 24 = -14x. We could, of course, add 14x to both sides and factor, but this only works if the equation does factor. Not all will. This is why we now will learn the Completing the Square method. This method makes use of the following formula:
(x + y)2 = x2 + 2xy + y2
You can verify this formula with the FOIL method, or there is a beautiful geometric explanation that involves a big square, a little square, and two rectangles. I will not cover these here, but I will write a future post on multiplying polynomials that covers this in detail. For now, I ask that you either verify it for yourself or take my word for it.
To do the Completing the Square method, first rearrange the terms so that the variable terms are all on one side and the constant term is on the other side. Then, if the leading coefficient is not 1, divide by that number on both sides:
4x2 + 24 = -28x
4x2 + 28x = -24
x2 + 7x = -6
To use the equations above, the coefficient of the middle term will be 2y. Take half of this to find y, and then add y2 to both sides of the equation.
y = 7/2. Add (7/2)2 to both sides of the equation.
x2 + 7x = -6
x2 + 7x + (7/2)2 = -6 + (7/2)2
(x + 7/2)2 = -24/4 + 49/4
(x + 7/2)2 = 25/4
x + 7/2 = ±5/2
x = -7/2 ± 5/2
x = -7/2 + 5/2 or x = -7/2 - 5/2
x = -1, -6
Check: x = -1
4(-1)2 + 24 = -28(-1)?
4 + 24 = 28? YES
Check: x = -6
4(-6)2 + 24 = -28(-6)?
144 + 24 = 168? YES
Now that we have been through the process once, lets boil down the actual mechanics.
- Rearrange the terms if needed
- Divide by the leading coefficient
- Add (half the coefficient of x)2 to both sides
- Write the variable side as a perfect square
- Take the square root of both sides
- Solve the linear equations
Lets try this on another:
9x2 - 29x + 6 = 0
9x2 - 29x = -6
x2 - 29/9 x = -2/3
x2 - 29/9 x + (-29/18)2 = -2/3 + 841/324
(x - 29/18)2 = -216/324 + 841/324
(x - 29/18)2 = 625/324
x - 29/18 = ±25/18
x = 29/18 ± 25/18
x = 54/18 or x = 4/18
x = 3 or x = 2/9
Check: x = 3
9(3)2 - 29(3) + 6 = 0?
81 - 87 + 6 = 0? YES
Check: x = 2/9
9(2/9)2 - 29(2/9) + 6 = 0?
4/9 - 58/9 + 54/9 = 0? YES
Granted, this was tedious and could have been solved faster factoring by the AC method, but it is a good example to see that even when there are fractions involved the method still works. The final two examples in this section will be quadratic equations that do not have rational solutions:
x2 - 5x + 3 = 0
x2 - 5x = -3
x2 - 5x + (-5/2)2 = -3 + 25/4
(x - 5/2)2 = -12/4 + 25/4
(x - 5/2)2 = 13/4
x - 5/2 = ± √(13/4)
x - 5/2 = ± √(13)/2
x = 5/2 ± √(13)/2 * My preferred solution expression
x = (5 ± √(13))/2 * Another popular solution expression
Check: 5/2 + √(13)/2
(5/2 + √(13)/2)2 - 5(5/2 + √(13)/2) + 3 = 0?
25/4 + 5√(13)/2 + 13/4 - 25/2 - 5√(13)/2 + 3 = 0?
38/4 - 25/2 + 3 = 0?
19/9 - 25/2 + 3 = 0?
-6/2 + 3 = 0? YES
Check: 5/2 - √(13)/2
(5/2 - √(13)/2)2 - 5(5/2 - √(13)/2) + 3 = 0?
25/4 - 5√(13)/2 + 13/4 - 25/2 + 5√(13)/2 + 3 = 0?
38/4 - 25/2 + 3 = 0?
19/9 - 25/2 + 3 = 0?
-6/2 + 3 = 0? YES
Final Completing the square example:
x2 + 6x + 11 = 0
x2 + 6x = -11
x2 + 6x + (3)2 = -11 + 9
(x + 3)2 = -2
x + 3 = √(-2)
The square root of a negative number is not a real number. It is a complex number, and I will deal with complex numbers in a future post. For the purposes of this post, any time I get a square root of a negative I will stop and report the answer as “No real solutions”
x2 = 121
x2 - 121 = 0
(x - 11)(x + 11) = 0
x = 11, or -11
Another way to write this is x = ±11, read “x equals plus or minus 11”. Checking is very easy because 112 = 121 and (-11) 2 = 121. As you can see, 11 was only part of the solution. This is because squaring has the nice property that both negative and positive values turn out positive when the value is squared. Thus, whenever you take the square root of both sides of an equation you have to include both the positive and negative values so you must always introduce the ± in your answer or write two separate equations. To solve this equation without factoring you would write it as follows:
x2 = 121
√(x2) = ±√(121)
x = ±11
Now lets do another one:
(2x + 7)2 = 25
2x + 7 = ±5
2x = -7 ± 5
x = (-7 ± 5)/2
x = (-7 + 5)/2 or x = (-7 - 5)/2
x = -1, -6
Check: x = -1
(2(-1) + 7)2 = 25?
52 = 25? YES
Check: x = -6
(2(-6) + 7)2 = 25?
(-5)2 = 25? YES
So that is not so bad as long as the quadratic is written as a perfect square. But now we need to figure out how to solve an equation like 4x2 + 24 = -14x. We could, of course, add 14x to both sides and factor, but this only works if the equation does factor. Not all will. This is why we now will learn the Completing the Square method. This method makes use of the following formula:
(x + y)2 = x2 + 2xy + y2
You can verify this formula with the FOIL method, or there is a beautiful geometric explanation that involves a big square, a little square, and two rectangles. I will not cover these here, but I will write a future post on multiplying polynomials that covers this in detail. For now, I ask that you either verify it for yourself or take my word for it.
To do the Completing the Square method, first rearrange the terms so that the variable terms are all on one side and the constant term is on the other side. Then, if the leading coefficient is not 1, divide by that number on both sides:
4x2 + 24 = -28x
4x2 + 28x = -24
x2 + 7x = -6
To use the equations above, the coefficient of the middle term will be 2y. Take half of this to find y, and then add y2 to both sides of the equation.
y = 7/2. Add (7/2)2 to both sides of the equation.
x2 + 7x = -6
x2 + 7x + (7/2)2 = -6 + (7/2)2
(x + 7/2)2 = -24/4 + 49/4
(x + 7/2)2 = 25/4
x + 7/2 = ±5/2
x = -7/2 ± 5/2
x = -7/2 + 5/2 or x = -7/2 - 5/2
x = -1, -6
Check: x = -1
4(-1)2 + 24 = -28(-1)?
4 + 24 = 28? YES
Check: x = -6
4(-6)2 + 24 = -28(-6)?
144 + 24 = 168? YES
Now that we have been through the process once, lets boil down the actual mechanics.
- Rearrange the terms if needed
- Divide by the leading coefficient
- Add (half the coefficient of x)2 to both sides
- Write the variable side as a perfect square
- Take the square root of both sides
- Solve the linear equations
Lets try this on another:
9x2 - 29x + 6 = 0
9x2 - 29x = -6
x2 - 29/9 x = -2/3
x2 - 29/9 x + (-29/18)2 = -2/3 + 841/324
(x - 29/18)2 = -216/324 + 841/324
(x - 29/18)2 = 625/324
x - 29/18 = ±25/18
x = 29/18 ± 25/18
x = 54/18 or x = 4/18
x = 3 or x = 2/9
Check: x = 3
9(3)2 - 29(3) + 6 = 0?
81 - 87 + 6 = 0? YES
Check: x = 2/9
9(2/9)2 - 29(2/9) + 6 = 0?
4/9 - 58/9 + 54/9 = 0? YES
Granted, this was tedious and could have been solved faster factoring by the AC method, but it is a good example to see that even when there are fractions involved the method still works. The final two examples in this section will be quadratic equations that do not have rational solutions:
x2 - 5x + 3 = 0
x2 - 5x = -3
x2 - 5x + (-5/2)2 = -3 + 25/4
(x - 5/2)2 = -12/4 + 25/4
(x - 5/2)2 = 13/4
x - 5/2 = ± √(13/4)
x - 5/2 = ± √(13)/2
x = 5/2 ± √(13)/2 * My preferred solution expression
x = (5 ± √(13))/2 * Another popular solution expression
Check: 5/2 + √(13)/2
(5/2 + √(13)/2)2 - 5(5/2 + √(13)/2) + 3 = 0?
25/4 + 5√(13)/2 + 13/4 - 25/2 - 5√(13)/2 + 3 = 0?
38/4 - 25/2 + 3 = 0?
19/9 - 25/2 + 3 = 0?
-6/2 + 3 = 0? YES
Check: 5/2 - √(13)/2
(5/2 - √(13)/2)2 - 5(5/2 - √(13)/2) + 3 = 0?
25/4 - 5√(13)/2 + 13/4 - 25/2 + 5√(13)/2 + 3 = 0?
38/4 - 25/2 + 3 = 0?
19/9 - 25/2 + 3 = 0?
-6/2 + 3 = 0? YES
Final Completing the square example:
x2 + 6x + 11 = 0
x2 + 6x = -11
x2 + 6x + (3)2 = -11 + 9
(x + 3)2 = -2
x + 3 = √(-2)
The square root of a negative number is not a real number. It is a complex number, and I will deal with complex numbers in a future post. For the purposes of this post, any time I get a square root of a negative I will stop and report the answer as “No real solutions”
2. Quadratic Formula: Completing the square can be tedious and boring. It would be nice if we did not have to wade through all the steps and could just jump to the answer. This is exactly what the quadratic formula does. To derive the quadratic formula, start with a generic quadratic equation and solve it by completing the square:
ax2 + bx + c = 0
ax2 + bx = -c
x2 + b/a x = -c/a
x2 + b/a x + (b/(2a))2 = -c/a + b2/(4a2)
(x + b/(2a))2 = b2/(4a2) - 4ac/(4a2)
(x + b/(2a))2 = (b2 - 4ac)/(4a2)
x + b/(2a) = ±√((b2 - 4ac)/(4a2))
x + b/(2a) = ±√(b2 - 4ac)/(2a)
x = -b/(2a) ± √(b2 - 4ac)/(2a) * My preferred expression
x = (-b ± √(b2 - 4ac))/(2a) * Another popular expression
The quadratic formula, the way I like to write it, is x = -b/(2a) ± √(b2 - 4ac)/(2a). I prefer to write it as two separate fractions because I have seen students make fewer simplification mistakes when using it this way, and because it will be more informative when you learn other topics such as the axis of symmetry of a parabola (another future post). Most text books write it as a single fraction with -b ± √(b2 - 4ac all in the numerator and 2a in the denominator. Be aware that both are valid.
Now that we have the quadratic formula, what do we do with it? c For any quadratic equation, all you need to do to solve it is to first put it in standard form, ax2 + bx + c = 0, identify the coefficients a, b, and c, and use these values in the quadratic formula. As examples I will re-solve all the quadratic equations that I solved with completing the square (checking will not be doe since I already checked them all):
4x2 + 24 = -28x
4x2 + 28x + 24 = 0
a = 4, b = 28, c = 24
x = -28/(2*4) ± √(282 - 4*4*24)/(2*4)
x = -7/2 ± √(784 - 384)/8
x = -7/2 ± √(400)/8
x = -7/2 ± 20/8
x = -7/2 ± 5/2
x = -1, -6
9x2 - 29x + 6 = 0
a = 9, b = -29, c = 6
x = 29/(2*9) ± √((-29)2 - 4*9*6)/(2*9)
x = 29/18 ± √(841 - 216)/18
x = 29/18 ± √(625)/18
x = 29/18 ± 25/18
x = 3, 2/9
x2 - 5x + 3 = 0
a = 1, b = -5, c = 3
x = 5/(2*1) ± √((-5)2 - 4*1*3)/(2*1)
x = 5/2 ± √(25 - 12)/2
x = 5/2 ± √(13)/2
x2 + 6x + 11 = 0
a = 1, b = 6, c = 11
x = -6/(2*1) ± √(62 - 4*1*11)/(2*1)
x = -3 ± √(36 - 44)/2
x = -3 ± √(-8)/2
No real solutions
The quadratic formula can even be used to solve very simple quadratic equations:
x2 = 121
x2 - 121 = 0
a = 1, b = 0, c = -121
x = 0/(2*1) ± √(02 - 4*1*(-121))/(2*1)
x = ±√(484)/2
x = ±22/2
x = ±11
Another application of the quadratic formula is to factor a quadratic expression for which it is difficult for you to find the factors or even one that does not have rational factors. For example lets factor the expression 60x2 - 17x - 21. First I will set it equal to 0, factor out any common constant factor if any, and use the quadratic formula solve the equation. Next I will write each answer separately. Finally, I will rearrange the two answer equations to be equal to zero with no fractions. This will give the factors.
60x2 - 17x - 21 = 0
a = 60, b = -17, c = -21
x = 17/(2*60) ± √((-17)2 - 4*60*(-21))/(2*60)
x = 17/120 ± √(5329)/120
x = 17/120 ± 73/120
x = 90/120, x = -56/120
x = 3/4, x = -7/15
4x = 3, 15x = -7
4x - 3 = 0, 15x + 7 = 0
60x2 - 17x - 21 = (4x - 3)(15x + 7)
You can check this with the FOIL method. One final example will show how to write a quadratic expression in factored form even if the expression has no rational factors:
x2 + 2x - 4
x2 + 2x - 4 = 0
a = 1, b = 2, c = -4
x = -2/(2*1) ± √(22 - 4*1*(-4))/(2*1)
x = -1 ± √(20)/2
x = -1 ± 2√(5)/2
x = -1 ± √(5)
x = -1 + √(5), x = -1 - √(5)
x + 1 - √(5) = 0, x + 1 + √(5) = 0
x2 + 2x - 4 = (x + 1 - √(5))(x + 1 + √(5))
ax2 + bx + c = 0
ax2 + bx = -c
x2 + b/a x = -c/a
x2 + b/a x + (b/(2a))2 = -c/a + b2/(4a2)
(x + b/(2a))2 = b2/(4a2) - 4ac/(4a2)
(x + b/(2a))2 = (b2 - 4ac)/(4a2)
x + b/(2a) = ±√((b2 - 4ac)/(4a2))
x + b/(2a) = ±√(b2 - 4ac)/(2a)
x = -b/(2a) ± √(b2 - 4ac)/(2a) * My preferred expression
x = (-b ± √(b2 - 4ac))/(2a) * Another popular expression
The quadratic formula, the way I like to write it, is x = -b/(2a) ± √(b2 - 4ac)/(2a). I prefer to write it as two separate fractions because I have seen students make fewer simplification mistakes when using it this way, and because it will be more informative when you learn other topics such as the axis of symmetry of a parabola (another future post). Most text books write it as a single fraction with -b ± √(b2 - 4ac all in the numerator and 2a in the denominator. Be aware that both are valid.
Now that we have the quadratic formula, what do we do with it? c For any quadratic equation, all you need to do to solve it is to first put it in standard form, ax2 + bx + c = 0, identify the coefficients a, b, and c, and use these values in the quadratic formula. As examples I will re-solve all the quadratic equations that I solved with completing the square (checking will not be doe since I already checked them all):
4x2 + 24 = -28x
4x2 + 28x + 24 = 0
a = 4, b = 28, c = 24
x = -28/(2*4) ± √(282 - 4*4*24)/(2*4)
x = -7/2 ± √(784 - 384)/8
x = -7/2 ± √(400)/8
x = -7/2 ± 20/8
x = -7/2 ± 5/2
x = -1, -6
9x2 - 29x + 6 = 0
a = 9, b = -29, c = 6
x = 29/(2*9) ± √((-29)2 - 4*9*6)/(2*9)
x = 29/18 ± √(841 - 216)/18
x = 29/18 ± √(625)/18
x = 29/18 ± 25/18
x = 3, 2/9
x2 - 5x + 3 = 0
a = 1, b = -5, c = 3
x = 5/(2*1) ± √((-5)2 - 4*1*3)/(2*1)
x = 5/2 ± √(25 - 12)/2
x = 5/2 ± √(13)/2
x2 + 6x + 11 = 0
a = 1, b = 6, c = 11
x = -6/(2*1) ± √(62 - 4*1*11)/(2*1)
x = -3 ± √(36 - 44)/2
x = -3 ± √(-8)/2
No real solutions
The quadratic formula can even be used to solve very simple quadratic equations:
x2 = 121
x2 - 121 = 0
a = 1, b = 0, c = -121
x = 0/(2*1) ± √(02 - 4*1*(-121))/(2*1)
x = ±√(484)/2
x = ±22/2
x = ±11
Another application of the quadratic formula is to factor a quadratic expression for which it is difficult for you to find the factors or even one that does not have rational factors. For example lets factor the expression 60x2 - 17x - 21. First I will set it equal to 0, factor out any common constant factor if any, and use the quadratic formula solve the equation. Next I will write each answer separately. Finally, I will rearrange the two answer equations to be equal to zero with no fractions. This will give the factors.
60x2 - 17x - 21 = 0
a = 60, b = -17, c = -21
x = 17/(2*60) ± √((-17)2 - 4*60*(-21))/(2*60)
x = 17/120 ± √(5329)/120
x = 17/120 ± 73/120
x = 90/120, x = -56/120
x = 3/4, x = -7/15
4x = 3, 15x = -7
4x - 3 = 0, 15x + 7 = 0
60x2 - 17x - 21 = (4x - 3)(15x + 7)
You can check this with the FOIL method. One final example will show how to write a quadratic expression in factored form even if the expression has no rational factors:
x2 + 2x - 4
x2 + 2x - 4 = 0
a = 1, b = 2, c = -4
x = -2/(2*1) ± √(22 - 4*1*(-4))/(2*1)
x = -1 ± √(20)/2
x = -1 ± 2√(5)/2
x = -1 ± √(5)
x = -1 + √(5), x = -1 - √(5)
x + 1 - √(5) = 0, x + 1 + √(5) = 0
x2 + 2x - 4 = (x + 1 - √(5))(x + 1 + √(5))
In conclusion, I’d like to mention that although we only used the Completing the Square method as a stepping stone for deriving the Quadratic Formula, there are many future uses for completing the square such as finding important points for circles, parabolas, ellipses, and hyperbolas.
In my next post I will discuss solving various other kinds of equations, many of which will give rise to quadratic equations in the solutions. Some of these methods will also give rise to extraneous solutions, which means a solution that was introduced during the solving of the problem that is not really a solution of the original equation.
In my next post I will discuss solving various other kinds of equations, many of which will give rise to quadratic equations in the solutions. Some of these methods will also give rise to extraneous solutions, which means a solution that was introduced during the solving of the problem that is not really a solution of the original equation.
