While the ultimate goal is the same, to determine the value(s) that hold true for the equation, solving quadratic equations requires much more than simply isolating the variable, as is required in solving linear equations. This piece will outline the different types of quadratic equations, strategies for solving each type, as well as other methods of solutions such as Completing the Square and using the Quadratic Formula. Knowledge of factoring perfect square trinomials and simplifying radical expression is needed for this piece. Let’s take a look! Standard Form of a Quadratic Equation ax2+ bx+c=0
Where a, b, and c are integers and a?
I. To solve an equation in the form ax2+c=k, for some value k. This is the simplest quadratic equation to solve because the middle term is missing.

Strategy: To isolate the square term and then take the square root of both sides.
Ex. 1) Isolate the square term, divide both sides by 2 Take the square root of both sides 2×2=40 2×22= 40 2 x2 =20 Remember there are two possible solutions x2= 20 Simplify radical; Solutions x= ± 20 x=± 25 (Please refer to previous instructional materials Simplifying Radical Expressions )
II. To solve a quadratic equation arranged in the form ax2+ bx=0.
Strategy: To factor the binomial using the greatest common factor (GCF), set the monomial factor, and the binomial factor equal to zero, and solve.
Ex. 2) 12×2- 18x=0 6x2x-3= 0Factor using the GCF 6x=0 2x-3=0Set the monomial and binomial equal to zero x=0 x= 32Solutions. In some cases, the GCF is simply the variable with a coefficient of 1.
III. To solve an equation in the form ax2+ bx+c=0, where the trinomial is a perfect square. This too is a simple quadratic equation to solve, because it factors into the form m2=0, for some binomial m. For factoring instructional methods, select.
The Easy Way to Factor Trinomials
Strategy: To the factor, the trinomial, set each binomial equal to zero, and solve.
Ex. 3) x2+ 6x+9=0 x+32=0Factor as a perfect square x+3x+3= 0Not necessary, but valuable step to show two solutions x+3=0 x+3=0Set each binomial equal to zero x= -3 x= -3 Solve x= -3
Double root solution
IV. To solve an equation in the form ax2+ bx+c=0, where the trinomial is not a perfect square, but factorable. Similar to the last example, this is a simple quadratic equation to solve, because it factors into the form mn=0, for some binomials m and n.
Strategy: To the factor, the trinomial, set each binomial equal to zero, and solve.
Ex. 4) 2×2-x-6=0 * Using the factoring method from The Easy Way to Factor Trinomials, we need to find two numbers that multiply to give ac, or -12 and add to give b, or -1. These values are -4 and 3. Rewrite the trinomial with these two values as coefficients to x that add to the current middle term of -1x. 2×2- 4x+3x-6=0Rewrite middle term 2×2- 4x+3x-6=0 2xx-2+ 3x-2= 0Factor by grouping x-22x+3= 0Factor out the common binomial (x-2) x-2=0 2x+3=0Set each binomial equal to zero x=2 x= -32
Solutions
V. To solve a quadratic equation not arranged in the form ax2+ bx+c=0, but factorable.
Strategy: To
combine like terms to one side, set equal to zero, factor the trinomial set each binomial equal to zero, and solve. Ex. 5) 6×2+ 2x-3=9x+2 -9x -9x 6×2- 7x-3= 2 -2 -2 6×2- 7x-5=0 * To factor this trinomial, we are looking for two numbers that multiply to give ac, or -30, and add to give b, or -7. These values would be 3 and -10. Rewrite the trinomial with these two values as coefficients to x that add to the current middle term of -7x. 6×2+ 3x-10x-5=0. Rewrite middle term 6×2+ 3x-10x-5=0 3x2x+1-52x+1=0.
Factor by grouping
Careful factoring a -5 from the second group 2x+13x-5=0 Factor out the common binomial (2x+1) 2x+1=0 3x-5=0 Set each binomial equal to zero x= -12 x= 53
Solutions
Now that we have explored some examples, I’d like to take this time to summarize the strategies used thus far in solving quadratic equations. Keeping in mind the goal is to isolate the variable, the format of the equation will dictate the strategy used to solve it. When the quadratic does not have a middle term, a term with a power of 1, it is best to first isolate the squared term, and then take the square root of both sides. This essentially will result in two solutions of opposite values. For quadratics that do not have a c-value, arrange the equation so that ax2+ bx=0, and then factor using the GCF. Set the monomial, or the GCF, and the binomial equal to zero and solve. When the quadratic has one or more ax2’s, bx’s, and c’s, the like terms need to be combined to one side of the equation and set equal to zero before determining if the trinomial can be factored. Once factored, set each binomial equal to zero and solve. Keep in mind while combining like terms that a must be an integer greater than or equal to 1. The solutions to cases such as these may result in a double root solution, found when the trinomial is factored as a perfect square, or two unique solutions, found when the trinomial is factored into two unique binomials. There may be other cases where a GCF can be factored out of the trinomial before factoring occurs. Since this unit is focused on solving quadratic equations, the GCF would simply be a constant. The next example to illustrates while it’s helpful to factor out the GCF before factoring the trinomial, it is not imperative to do so and has no impact on the solution of the quadratic equation.
VI. To solve a quadratic equation in which there is a GCF among the terms of a trinomial.
Strategy A: To determine the GCF between the terms of the trinomial once it is in standard form, factor out the GCF, factor the trinomial, set each binomial equal to zero, and then solve. Ex. 6A) 12×2- 22x+6=0 26×2- 11x+3=0 * To factor this trinomial, we are looking for two numbers that multiply to give ac, or 18, and add to give b, or -11. These values would be -9 and -2. Rewrite the trinomial with these two values as coefficients to x that add to the current middle term of -11x. 26×2- 9x-2x+3=0Factor out the GCF of 2 from each term 3x2x-3- 12x-3=0Factor by grouping 22x-33x-1=0Factor out the common binomial (2x-3) 2x-3=0 3x-1=0Set each binomial equal to zero x= 32 x= 13.
Solutions Strategy (B): To the factor, the trinomial, set each binomial equal to zero, and solve. Ex. 6B) 12×2- 22x+6=0. To factor this trinomial, we are looking for two numbers that multiply to give ac, or 72, and add to give b, or -22. These values would be -18 and -4. Rewrite the trinomial with these two values as coefficients to x that add to the current middle term of -22x. 12×2- 18x-4x+6=0 x2x-3- 22x-3=0. Factor by grouping 2x-36x-2= 0Factor out the common binomial (2x-3) 2x-3=0 6x-2=0 Set each binomial equal to zero x= 32 x= 26= 13Solutions. Notice in Ex 6A, since the GCF did not have a variable. The purpose of factoring and setting each binomial equal to zero is to solve for the possible value(s) for the variable that results in a zero product. If the GCF does not have a variable, it is not possible for it to make a product of zero. With that said, in later topics there will be cases where a GCF will include a variable, leaving a factorable trinomial.
This type of case results in a possibility of three solutions for the variable, as seen in the example below. 3xx2+ 5x+6=0 3xx+2x+3=0 3x=0 x+2=0 x+3=0 x=0 x= -2 x= -3. At this point, we need to transition to solving quadratics equations that do not have trinomials that are factorable. To solve these types of equations, we have two options, (1) to Complete the Square, and (2) to use the Quadratic Formula. Essentially, these two methods yield the same solution when left in simplified radical form.
For the remainder of this unit I will o the following:

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Explain how to Complete the Square
Provide examples utilizing the Completing the Square method
Prove the Quadratic Formula starting with Completing the Square
Provide examples solving equations using the Quadratic Formula
Provide an example that parallels all three methods in this unit
Provide instructional strategies for solving quadratic equations

VII. How to Complete the Square Goal: To get x±m2=k, where m and k are real numbers and k? 0 For equations that are not factorable and in the form ax2+ bx+c=0 where a=1,

Move constant term to the side opposite the variable x.
Take 12 of b and square the result.
Add this term to both sides.
Create your perfect square set equal to some constant value k? 0.

VIII. To solve quadratic equations using the Completing the Square method.
Ex. 7) x2+ 6x-5=0 Since there are no two integers that multiply to give ac, or -5, and add to give b, or 6, this trinomial is not factorable, and therefore, Completing the Square must be used to solve for x. x2+ 6x+ _____ =5+ _____ Move constant to the right x2+ 6x+ 62 2=5+ 62 2Take 12b, square it and add it to both sides 2+ 6x+9=14. Simplify x+32=14. Factor trinomial as a perfect square x+32= 14. Take the square root of both sides x+3= ± 14. Simplify x= -3 ± 14Solve for x;
Solutions
Ex. 8) 2×2+ 16x=4. Before proceeding with Completing the Square, notice a? 1, and the constant term is already on the opposite side of the variable terms. The first step must be to divide both sides of the equation by 2. x2+ 8x=2. Result after division by 2 x2+ 8x+ _____ =2+ _____.
Preparation for Completing the Square
x2+ 8x+ 82 2=2 + 82 2. Take 12b, square it and add it to both sides x2+ 8x+16=18. Simplify x+42=18. Factor trinomial as a perfect square +42= 18. Take the square root of both sides x+4= ± 32. Simplify x= -4 ±32. Solve for x; Solutions At any point during the solving process, if a negative value exists under the radical, there will be NO REAL SOLUTION to the equation. These types of equations will be explored later once the imaginary number system has been learned.
The Quadratic Formula is another method for solving a quadratic equation. Let’s take a look at how the standard form of a quadratic equation can be transformed into the Quadratic Formula Using the Completing the Square method.
Ensure a coefficient of 1 for x2 by dividing by a, and move the constant term to the right ax2+ bx+c=0
Standard Form of a quadratic equation ax2a+ bxa+ c a= 0 a x2+ b ax+ c a= 0 x2+ b ax= – c

a.  The square of half of what is now the b term, or the middle term, is 12
b. a2= b2a2= b24a2 Complete the Square Get common denominator on the right Factor trinomial as a perfect square

Take the square root of both sides Simplify Solve for x Quadratic Formula x2+ b ax+ b24a2 = – c a + b24a2 x2+ b ax+ b24a2 = – 4ac 4a2 + b24a2 x2+ b ax+ b24a2 = -4ac+b24a2 + b 2a2= -4ac+b24a2 x+ b 2a2= -4ac+b24a2 x+ b2a= ± -4ac+ b22a x= -b 2a ± -4ac+ b22a x= -b ± b2- 4ac2a X. To solve quadratic equations using the Quadratic Formula.
Ex 9.) 2×2- 8x+ 5=0 a=2 b= -8 c=5 Substitute Evaluate Subtract Simplify radical Simplify fraction; Solutions x= -b ± b2- 4ac2a x= –8 ± -82- 42522 x= 8 ± 64 – 404 x= 8 ± 244 x= 8 ± 264 x= 4 ± 62
Ex. 10) 2x=5-4×2 * Notice this equation is not in the standard form for quadratic equations. Before identifying the values for a, b and c, the equation must be arranged in ax2+ bx+c=0 form.
After adding 4×2 and subtracting 5, we get 4×2+ 2x-5=0 a=4 b= 2 c=-5 Substitute Evaluate Add Simplify x= -b ± b2- 4ac2a x= -2 ± 22- 44-524 x= -2 ± 4+808 x= -2 ± 848 x= -2 ± 2218 Simplify fraction; Solution x= -1 ± 214 As in Completing the Square, if a negative value results under the radical, there’s NO REAL SOLUTION.
XI. Compare all three methods learned Factoring
Ex. 11) 4×2- 8x-5=0 * Two integers that multiply to give -20 that add to give -8 are -10 and 2. x2- 10x +2x-5=02x2x-5+ 12x-5= 02x-52x+1= 0 2x-5=0 2x+1=0x= 52 x= -12| Ex. 11) 4×2- 8x-5=0 * First step is to obtain a coefficient of 1 for the x2 by dividing both sides of the equation by 4. x2- 2x- 54= 04×2- 2x- 54=0x2- 2x=54×2- 2x+ _____=54+ _____x2-2x+ 22 2 =54+ 22 2 x2- 2x+1=54+1×2- 2x+1=94x-12= 94x-12= 94x-1= ± 32x=1 ± 32x= 52 x= -12| Ex. 11. ) 4×2- 8x-5=0a=4 b= -8 c= -5x= -b ± b2- 4ac2ax= –8 ± (-8)2-44-52(4)x= 8 ± 64+80 8x=8 ± 1448x= 8 ± 128x= 208 x= -48x= 52 x= -12
XII. Instructional Strategies
Often students will gravitate towards the formula because they are comfortable with mindless substitution and computation that’s involved with a formula. Needless to say, they quickly realize they must be meticulous weaving in and out of the steps so as not to lose a sign or simplify incorrectly. In many cases, taking the scenic route, or the more elaborate method of the solution will cause careless errors throughout the solving process. The goal is for students to learn the process of examining what they have been given and proceed with the method of solution that makes sense for the given equation. To encourage this type of analysis and discourse, provide opportunities for students to showcase these skills. One activity is to group students in 3’s, provide them with a quadratic equation to solve, have each student demonstrate one of the methods of solution, and then decide as a group which method was the most efficient or strategic. When presenting to the class, have each student explain why their method was or was not the most efficient. In a class, this could be 10 or more equations solved. Don’t shy away from including equations that are missing terms or equations that are not in standard form. These might prove to be more difficult, since they are required to think more carefully about what they have been given, but they are very valuable learning tools. Following this activity, provide students with an equation, and without requiring them to solve using paper and pencil, have them explain, either verbally or in written form, which method they think would be the most strategic or most efficient. Keep in mind, there is room for opinion in these responses. Simply listen and evaluate students’ thought processes as they explain. Skills such as these are invaluable and will help create well rounded mathematical thinkers.

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