We will examine three methods of solving quadratic equations, factoring, the completion of the square, and the quadratic formula. Sometimes a technique like factoring, has limited use. While the quadratic formula always works, it may not be the best choice, and does not make the underlying idea clear. It is necessary to study all three methods in order to properly comprehend how we solve quadratic equations.
The Retired Professor Simplifies Algebra: Quadratic Equations
This simple understanding of algebra, the quadratic equation, is intended as a free resource for students and parents of students to help them grasp the ideas of algebra.
A quadratic equation is an equation that can be put in the form of a second degree polynomial equal to zero. A second-degree polynomial has an x2 term, and possibly an x term, and perhaps a constant term, but no other appearances of the variable. Sometimes terms have to be moved from the side of the equation that will become zero by addition or subtraction. Sometimes symbols of grouping will need to be removed, a process often requiring the distribution of multiplication over addition. The distribution property is a(b + c) = ab + ac. And sometimes like terms, those to with the variable raised to the same power, will need to be combined.
A quadratic equation can have as many as two solutions. Some have only one solution, which is repeated. It is possible the solutions are complex numbers. If they are, the solutions are conjugates of each other, meaning if one is of the form a + bi the other is of the form a – bi.
The Methods of Solving Quadratic Equations
Needed Other Ideas
The Retired Professor Simplifies Algebra: FactoringThis algebra insight is intended to assist students and those helping students with the learning process. It is a free to use resource. This snipet covers factoring.
Factoring
First, setting the quadratic equation equal to zero allows use of a powerful tool. If two numbers are multiplied together, and that product is equal to zero, at least one of those numbers equals zero.
If we think of factors as numbers, we can set each factor equal to zero, solve the two linear equations produced, and find the solutions of the quadratic equation.
Solve x2 + 6x = -8.
Add 8 to each side to get x2 + 6x + 8 = 0.
Next, factor to get (x + 4)(x + 2) = 0
Set each factor equal to zero, and solve the linear equations.
x + 4 = 0 x = -4
x + 2 = 0 x = -2
So, the solutions for the quadratic equation are -4 and -2.
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Another technique one can use is completion of the square. This is helpful in other problems, so develop the technique.
Since not every equation can be factored there is a need for techniques other than factoring to use in such cases.
In factoring we discover two special forms of perfect squares.
x2 + 2bx + b2 = (x + b)2
x2 - 2bx + b2 = (x - b)2
Our objective is to get one of these forms.
Solve x2 + 6x – 7 = 0
First, move the constant to the other side of the equation by adding the same value to both sides. In this case adding 7 to each side is needed.
x2 + 6x = 7
Next, find b by dividing he coefficient of x by 2.
6/2 = 3
Now, square b and add b2 to each side of the equation.
x2 + 6x + 9 = 7 + 9
(x + 3)2 = 16
Now, take the square root of each side. It is allowed.
x + 3 = 4 and x + 3 = -4
There are two roots that occur when taking the square root.
Next, add -3 to each side of each equation.
x = 1 and x = -7
So, 1 and -7 are the solutions of the quadratic equation.
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Quadratic Formula
The quadratic formula is found by solving by the completion of the square for a general equation and inserting the values when the formula is used.
Solve ax2 + bx + c = 0
Add -c to each side.
ax2 + bx = -c
Next, divide by a.
x2 + bx/a = -c/a
Take half of the coefficient of x and divide by 2.
b/2a
Square the result.
b2/(4a)
This is what is needed to be added to each side of the equation.
x2 + bx/a + b2/(4a2) = b2/(4a2) – c/a
Make the right side of the equation a single fraction with the common denominator of 4a2.
x2 + bx/a + b2/(4a2) = (b2 –4ac)/(4a2)
The left side of the equation is a perfect square.
(x + b/2a)2 = (b2 –4ac)/(4a2)
Take the square root of each side.
(x + b/2a) = ±(b2 –4ac)1/2/(2a)
So, x = (-b ±(b2 –4ac)1/2)/(2a)
So, to solve a quadratic equation we get zero on one side, then use the quadratic formula.
Solve x2 + 6x = 7 using the quadratic formula.
Subtract 7 from each side of the equation.
x2 + 6x – 7 = 0
Identify a, b, and c: a = 1, b = 6, and c = -7.
x = (-6 ± (62 – 4(1)(-7)1/2)/(2x1)
x = (-6 ± 641/2)/2
x = 1 and x = -7 are the solutions.
It is important to know that when the value under the radical symbol is positive there are two real solutions, when it is zero there is one real repeated solution, and when it is negative there are two complex number solutions that are conjugates of each other.
The quadratic formula is nothing more than completing the square for a general equation. Once it is done, it can just be used thereafter. Or, you can just accept it and plug in.
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Comments
blackspanielgalleryFeb 12, 2021
DerdriuMarrinerFeb 12, 2021
Which of the three methods do you prefer? Where does completing the square fit between the limited use of factoring (my favorite because the most fun, albeit not always the most applicable) and the workability of the quadratic formula despite the latter's not always making "the underlying idea clear"?