Understanding Quadratic Equations

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The first thing one must know to understand quadratic equations is the term "quadratic". A quadratic equation is an algebraic equation in which the highest power of the variable is 2. For example x^2+2x+3=0 is a quadratic equation, whereas 2x+4=0, x^3+2x^2+x+5=0 are not. Generally speaking the standard form of a quadratic equation is ax^2+bx+c=0, where a,b, c are integers. To solve quadratic equations there are several methods one can use. I will discuss these methods below.

Suppose you wish to multiply (x + 5) and (x - 3). This is done by a process known as FOIL.  Multiply the first term in the set of parentheses, (x)(x). Then multiply the outer term in each set of parentheses, (x)(-3). Then multiply the inner terms (5)(x). Finally, multiply the last terms (5)(-3). When combining like terms we get x^2 -3x + 5x -15 = x^2 + 2x -15.  Suppose now that we wish to take the quadratic equation, x^2 + 2x - 15 and factor it. The factored form is (x + 5)(x - 3).  So the process I am going to introduce is basically a reverse FOIL. 

You will set up two sets of parentheses ( )( ). In each set you will have 2 terms, with an addition or subtraction sign between each term. First you have to think of what  two terms multiplied together gives the first term in the quadratic equation. That would be x and x, since x times x is x^2. Put them in the first location in each parenthesis. So you have (x    )(x     ). Next you want to see what multiplies together to give you the last term of -15, but ADDS to give you the middle term coefficient of 2.  That would be 5 and -3 since 5 times -3 equals -15 and 5 plus -3 equals 2.. Those numbers go in the remaining two spots in the parenthesis to give you (x + 5)(x- 3).

Suppose the quadratic equation used in the example above is set equal to zero. How do we solve such an equation? We simply factor it and set each factor equal to zero and solve. For example, (x + 5)(x - 3) = 0, so we have 2 equations to solve:

(x + 5) = 0 and (x - 3) = 0

Solving each equation gives us x = -5 or x = 3.

Note that in some quadratic equations, you must factor out the greatest common factor first to make factoring easier.  For example, if the equation is 4x^2 + 20x + 16, it's easier to factor out a 4 first to get 4(x^2 + 5x + 4).  From here, we can factor x^2 + 5x + 4 like we did above. The factored form is 4(x + 4)(x + 1).

The next way to factor is a method called completing the square.  The idea is when factoring to get a perfect square trinomial on the left side of the equation equaling a number on the right side and taking square root of both sides to solve for x.  A perfect square trinomial is a trinomial that is factored into a single term squared. For example x^2 + 6x + 9 is a perfect square trinomial because it is factored to (x +3)(x+ 3).

Suppose we have the problem x^2 + 4x + 3 = 0. 

First thing we do is make sure the coefficient (number in front of the variable) of the squared term is 1, if it isn't you must divide by that coefficient to get that to be 1.

Subtract 3 from both sides of the equation to get

x^2 + 4x = -3

We want to turn x^2 + 4x into a perfect square trinomial. To do that we take half the middle term coefficient, square it and add to both sides. This gives us

x^2 + 4x + 4 = -3 + 4

Notice now that the left side of the equation is a perfect square trinomial.  Factor this and add terms on the right side to get

(x + 2)(x + 2) = 1 or (x + 2)^2 = 1

Now we take the square root of both sides to eliminate the exponent of 2 from the left side of the equation.

(x  + 2) = +/-1

When we solve this for x, we get x = -3 and x = -1.

Generally speaking if you can factor by the reverse FOIL method, do so because this is a bit more complex, as is the next method.

The final method used to factor quadratic equations is by the quadratic formula. The basis of this is using the standard form of ax^2+bx+c=0 and using the a,b,c in a formula to solve for x. The formula is x=-b +/ - square root of b^2-4ac, all divided by 2a. In our example of x^2+4x+3=0, a is 1, b is 4 and c is 3. By substituting those numbers into that formula you get

x = -[4 +/- square root(4^2 - 4(1)(3))]/2(1)

x =[ -4 +/- square root(16 - 12)]/2

x =[ -4 +/- square root(4)]/2

x = (-4 +/- 2)/2

x = (-4 + 2)/2 or (-4 - 2)/2

x = -1 or x = -3.

Notice that all 3 methods give the same answer. You can use any of the methods you wish to factor a quadratic equation. This example has solutions for x that are real numbers and 2 answers. There is also a possibility that there is only 1 solution when there is a square, ( (x+2)^2=0, would give x= -2, which is called a double root) . You could also get an imaginary number as the solution if the square root part of the quadratic formula is negative.

The graphs of quadratic equations such as y = ax^2 + bx + c are parabolas which open upward. If a is negative it opens downward. IF the equation is in the form x = ay^2 + by + c, then the parabola opens left. If a is negative, the parabola opens to the left.  One can graph quadratic equations by setting up an x, y  chart and choose a few numbers for x, solve for y and plot on a rectangular coordinate system.  If the equation is y = ax^2 + bx + c, it is wise to use 0 for y and find the x intercepts. The vertex is then halfway between the intercepts. For example, if the intercepts are where the x coordinates are 0 and 4, you know the vertex is where x = 2.   The axis of symmetry for a parabola opening up or down is always the line the goes vertically through the vertex.

Likewise, graphing a parabola in the form x = ay^2 + by + c is easy if you find the y-intercepts first. This is done by putting 0 in for x and solving for y. The vertex will be halfway in between the intercepts and the axis of symmetry is the horizontal line through the vertex.

I have tutored algebra for 12 years and these are the methods I teach students all the time when trying to factor quadratic equations. I hope anyone that needs help has gotten a better understanding how to do this with my article.

More about this author: Kerry Kauffman

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