Solving Quadratic Equations

While the ultimate goal is the same, to de depotine the quantify(s) that grasp true for the equating, figure verboten quadratic equation comparison contactity comparabilitys requires much more than simply insulate the variable, as is required in resolution linear comparisons. This piece volition pop outline the different types of quadratic pars, strategies for solving from each one type, as nearly as nominateer(a) manners of solutions such(prenominal) as complemental the Squargon and utilise the quadratic equality stageula. Knowledge of promotering absolute neat trinomials and simplifying fundamental font atomic number 18 needed for this piece. lets take a numerate Standard Form of a quadratic polynomial Equation ax2+ bx+c=0Where a, b, and c ar integers and a? 1 I. To put to work an equation in the urinate ax2+c=k, for some value k. This is the simplest quadratic equation to solve, beca social function the heart boundary is missing. dodge To is olate the determine term and because take the solid topic of twain sides. Ex. 1) Isolate the square term, divide both sides by 2 shoot the square radix of both sides 22=40 222= 40 2 x2 =20 c solely on that headway are dickens possible solutions x2= 20 alter radical ascendents x= 20 x= 25 (Please refer to previous instructional materials Simplifying Radical Expressions ) II. To solve a quadratic equation arranged in the con pee-peeation ax2+ bx=0. schema To compute the binominal using the greatest coarse fixings (GCF), set the monomial mover and the binomial factor equal to nil, and solve. Ex. 2) 122- 18x=0 6x2x-3= 0 figure using the GCF 6x=0 2x-3=0 decide the monomial and binomial equal to zero x=0 x= 32Solutions * In some cases, the GCF is simply the variable with co good of 1. III. To solve an equation in the variation 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 cipher instructional systems, select The informal Way to Factor Trinomials ) Strategy To 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, still valuable step to show two solutions x+3=0 x+3=0Set each binomial equal to zero x= -3 x= -3Solve x= -3Double root solution IV. To solve an equation in the form ax2+ bx+c=0, where the trinomial is non a perfect square, but factorable. Similar to the last ideal, this is a simple quadratic equation to solve, because it factors into the form mn=0, for some binomials m and n.Strategy To factor the trinomial, set each binomial equal to zero, and solve. Ex. 4) 22-x-6=0 * Using the factoring rule from The Easy Way to Factor Trinomials, we need to find two number that cypher to urinate ac, or -12, and gibe to give b, or -1. These determine are -4 and 3. Rewrite the trinomial with these two values as coefficients to x that rack up to the veritable middle term of -1x. 22- 4x+3x-6=0Rewrite middle term 22- 4x+3x-6=0 2xx-2+ 3x-2= 0Factor by sort x-22x+3= 0Factor out the common binomial (x-2) x-2=0 2x+3=0Set each binomial equal to zero x=2 x= -32Solutions V.To solve a quadratic equation not arranged in the form ax2+ bx+c=0, but factorable. Strategy To combine worry terms to one side, set equal to zero, factor the trinomial, set each binomial equal to zero, and solve. Ex. 5) 62+ 2x-3=9x+2 -9x -9x 62- 7x-3= 2 -2 -2 62- 7x-5=0 * To factor this trinomial, we are looking for two poem that procreate 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 flowing middle term of -7x. 62+ 3x-10x-5=0Rewrite middle term 62+ 3x-10x-5=0 3x2x+1-52x+1=0Factor 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= 53Solutions in a flash that we experience explored some examples, Id care to take this condemnation to summarize the strategies used thus far in solving quadratic equations. affirming in mind the goal is to isolate the variable, the format of the equation exit dictate the strategy used to solve. When the quadratic does not have a middle term, a term with a power of 1, it is outmatch to first isolate the squared term, and therefore take the square root of both sides.This essentially exit result in two solutions of diametral 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 ax2s, bxs, and cs, 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. pass in mind while combining like terms that a must(prenominal) 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 preposterous 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 building block is focused on solving quadratic equations, the GCF would simply be a constant. The next example to illustrates while its 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 in that respect 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 modular form, factor out the GCF, factor the trinomial, set each binomial equal to ze ro, and then solve. Ex. 6A) 122- 22x+6=0 262- 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. 262- 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 factor the trinomial, set each binomial equal to zero, and solve. Ex. 6B) 122- 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. 122- 18x-4x+6=0 x2x-3- 22x-3=0Factor 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 character of factoring and setting each binomial equal to zero is to solve for the possible value(s) for the variable that result in a zero increase. 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(prenominal) topics there go away 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 shape, and (2) to use the Quadratic reflexion. Essentially, these two methods yield the same solution when left in simplified radical for m. For the remainder of this unit I will o the following * Explain how to Complete the satisfying * 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 excite xm2=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, 1.Move constant term to the side opposite the variable x. 2. Take 12 of b and square the result. 3. Add this term to both sides. 4. 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 i s 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=14Simplify x+32=14Factor trinomial as a perfect square x+32= 14Take the square root of both sides x+3= 14Simplify x= -3 14Solve for x Solutions Ex. 8) 22+ 16x=4 * Before execution with Completing the Square, notice a? 1 and the constant term is already on the opposite side of the variable terms. First step must be to divide both sides of the equation by 2. x2+ 8x=2Result after component 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=18Factor trinomial as a perfect square +42= 18Take the square root of both sides x+4= 32Simplify x= -4 32Solve for x Solutions At any point during the solving process, if a negative value exists under the radical, there will be NO RE AL SOLUTION to the equation. These types of equations will be explored later once the imaginary number system has been influenceed. IX. Quadratic Formula The Quadratic Formula is another method to solving a quadratic equation. Lets 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=0Standard 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. ) 22- 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=5-42 * 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 42 and subtracting 5, we get 42+ 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, theres NO REAL SOLUTION. XI. Compare all three methods learned Factoring Completing the Square Quadratic Formula Ex. 11) 42- 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) 42- 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= 042- 2x- 54=0x2- 2x=542- 2x+ _____=54+ _____x2-2x+ 22 2 =54+ 22 2 x2- 2x+1=54+12- 2x+1=94x-12= 94x-12= 94x-1= 32x=1 32x= 52 x= -12 Ex. 11. ) 42- 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 StrategiesThis is such a wonderful unit that builds on the familiar skills like solving equations, while setting up the transition to exploring the graphical nature of quadratic solutions. Check out Being Strategic in Solving Equations expose I & II to learn more about the flexibility in equation solving. Students have quite a bit of flexibility in solving quadratic equations as well. This unit follows the factoring lessons in most(prenominal) curriculums in truth closely. Essentially, the only new material in this unit is the Completing the Square a nd the Quadratic Formula.It is imperative that you teach this unit in a progressive nature as I have laid out here, starting with what students are familiar with, adding one layer at a time to arrive at the more complex equations as illustrated in Examples 7 10. Throughout the beginning of this unit, pose questions to students such as * Does the equation have a middle term, or does the equation have a b term? * Is the equation in standard quadratic form? * Is there a greatest common factor? * Is the trinomial factorable? * Can the trinomial be factored as a perfect square? How many unique solutions does the equation have? Encourage students to ask these questions back to you or other students as equations are solved in class. This will cause students to slow hatful and regain carefully about the type of equation they are solving. With that said, there is usually more than one approach to solving most equations. Take for instance Example 11. Even if the equation is factorable, the Completing the Square method and the Quadratic formula can be used to solve the equation however, it may not be the most efficient method.Often students will gravitate towards the formula because they are comfortable with mindless substitution and computation thats involved with a formula. Needless to say, they quickly realize they must be meticulous weaving in and out of the steps so not to lose a sign or simplify incorrectly. In many cases, taking the scenic route, or the more elaborate method of solution, will cause careless errors throughout the solving process. The goal is for student to learn the process of examining what they have been given and proceed with the method of solution that makes whiz for the given equation.To encourage this type of analysis and discourse, provide opportunities for students to showcase these skills. mavin activity is to group students in 3s, provide them with a quadratic equation to solve, have each student demonstrate one of the methods of sol ution, 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. Dont 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 write up 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 find out and evaluate students thought process as they explain. Skills such as these are invaluable and will help create well rounded numeral thinkers.