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Solving Quadratic Equations Solving Linear Equations β’ You know by now how to solve an equation such as 3π₯ β 19 = β4 β’ This equation can be written in the form ππ₯ + π = 0, and these are called linear equations β’ In fact, a linear equation is really just a first degree polynomial equation β’ Every first degree polynomial equation (or linear equation) can be solved using only the following 2 Solving Linear Equations β’ To solve 3π₯ β 19 = β4 you need β’ The properties of equality: π = π βΉ π + π = π + π and π = π βΉ ππ = ππ β’ The inverse property of addition: π + βπ = 0 1 π β’ The inverse property of multiplication: π β = 1, π β 0 β’ In this section you will learn methods for solving quadratic equations or second degree polynomial equations β’ A quadratic equation can be written in the form ππ₯ 2 + ππ₯ + π = 0, where π β 0 3 Solving Quadratic Equations β’ Recall that solving a linear equation involved using the required properties until you arrive at the equation 1 β π₯ + 0 = {solution} β’ You will need to arrive at the same equation in solving a quadratic equation β’ However, because quadratic equations involve an π₯ 2 term, the properties used to solve linear equations are not enough (though they are still valid and may be required) 4 Solving Quadratic Equations β’ To demonstrate, letβs use the three properties to try and solve the quadratic equation 5π₯ 2 β 1 = 124 β’ Add 1 to both sides to get 5π₯ 2 + 0 = 125 1 5 β’ Multiply both sides by (or divide by 5) to get 1 β π₯ 2 + 0 = 25 β’ As you can see, we donβt get 1 β π₯ + 0 = {solution}; we must do something about the π₯ 2 term 5 Solving Quadratic Equations β’ We run into a more difficult problem trying to solve the following quadratic equation π₯ 2 β 7π₯ β 12 = 0 β’ If you add 12 to both sides you get π₯ 2 β 7π₯ + 0 = 12 β’ But now, there is no way to combine the π₯ 2 term and the β7π₯ term since they are not like terms; youβre stuck after one step! 6 Solving Quadratic Equations β’ You will learn several techniques for solving quadratic equations β’ One technique will require that you use the definition of a square root that we developed earlier (somewhat modified) β’ Another technique will require you to factor β’ A third technique uses a formula called the quadratic formula β’ In general, every quadratic equation can be solved using the quadratic formula (if the equation has a solution) β’ But it is often easier to use the square root definition or factoring, so you will eventually have to choose the most efficient method 7 Solving Quadratic Equations β’ It will be helpful at the beginning to classify the types of quadratic equations by the best solution method β’ The general form of a quadratic equation is ππ₯ 2 + ππ₯ + π = 0 β’ The method you will use to solve an equation of this type depends on the values of π and π β’ If π = 0, the equation becomes ππ₯ 2 + π = 0; use the square root property (we will define this soon) β’ If π = 0, the equations becomes ππ₯ 2 + ππ₯ = 0 and this is easily solved by factoring β’ If neither π not π are zero, then either factor as a quadratic or use the quadratic formula 8 2 Solve Quadratic Equations: ππ₯ + π = 0 β’ Recall that our definition of square root was If π¦ 2 = π₯, then π¦ = π₯ β’ This is the definition of the principal square root (answer is always positive) β’ We need to modify this a bit to obtain the square root property β’ Notice that if π¦ 2 = 16, then π¦ = 4 is a solution since 42 = 16 β’ But it turns out that π¦ = β4 is also a solution because β4 2 = β4 β β 4 = 16 9 2 Solve Quadratic Equations: ππ₯ + π = 0 β’ So the first thing to remember about solving a quadratic equation of the form ππ₯ 2 + π = 0 is that there are always two solutions (unless the solution is zero) β’ Here, then, is the square root property (write it in your notes) β’ If π₯ 2 = π¦, then π₯ = π¦ or π₯ = β π¦ β’ We can see that this must be true because π¦ 2 = π¦ β π¦ = π¦2 = π¦ β’ And β π¦ 2 = β π¦ β β π¦ = π¦2 = π¦ 10 2 Solve Quadratic Equations: ππ₯ + π = 0 β’ The square root property is π₯ 2 = π¦ βΉ π₯ = π¦ or π₯ = β π¦ β’ Sometimes this is written more compactly as π₯2 = π¦ βΉ π₯ = ± π¦ β’ Now letβs use the square root property to solve the equation you saw earlier 5π₯ 2 β 1 = 125 β’ We were able to simplify this to π₯ 2 = 25 β’ Now by the square root property, π₯ = ± 25 = ±5 11 2 Solve Quadratic Equations: ππ₯ + π = 0 β’ Before practicing a few examples, notice that, although the general form is ππ₯ 2 + π = 0, this equation has solutions only if π β€ 0 β’ To see why, consider the equation π¦ = β1 β’ By the square root definition we must have π¦ 2 = β1, or π¦ β π¦ = β1 β’ But if π¦ = 1, then π¦ 2 = π¦ β π¦ = 1 β 1 = 1, not β1 β’ Also, if π¦ = β1, then π¦ 2 = π¦ β π¦ = β1 β β1 = 1, not β1 β’ There is no real number such that π¦ 2 = β1, and the same is true for any other negative numbers 12 2 Solve Quadratic Equations: ππ₯ + π = 0 β’ Now, if π > 0 in the equation π₯ 2 + π = 0, then we get π₯ 2 = βπ (by adding π to both sides of the equation) and this has no real number answer β’ So we should really think of the general form of the equation as ππ₯ 2 β π = 0 β’ Before we finish this chapter, we will solve the problem of β1 by inventing some new numbers 13 Guided Practice Use the square root property to solve each equation below. If it cannot be solved, write βno real number solution existsβ a) 2π₯ 2 β 7 = 25 b) β6π₯ 2 + 1 = β215 c) π₯ 2 β 18 = 0 d) 2π₯ 2 + 8 = 0 e) π₯ 2 = 98 f) 3π₯ 2 β 16 = β16 14 Guided Practice Use the square root property to solve each equation below. If it cannot be solved, write βno real number solution existsβ a) 2π₯ 2 β 7 = 25, π₯ = ±4 b) β6π₯ 2 + 1 = β215, π₯ = ±6 c) π₯ 2 β 18 = 0, π₯ = ± 18 = ±3 2 d) 2π₯ 2 + 8 = 0, no real number solution e) π₯ 2 = 98, π₯ = ± 98 = ±7 2 f) 3π₯ 2 β 16 = β16, π₯ = 0 15 2 Solving ππ₯ + ππ₯ + π = 0 by Factoring β’ Now we turn to solving quadratic equations of the form ππ₯ 2 + ππ₯ + π = 0 where neither π nor π is zero and where the expression can be factored β’ First, letβs review how to factor a quadratic expression β’ To factor 8π₯ 2 β 10π₯ + 3 β’ Multiply the lead coefficient by the constant to get 24 β’ Find the factors of 24 that add to β10: since β6 β β4 = 24 and β6 + β4 = β10, we use these β’ Rewrite the expression as 8π₯ 2 β 6π₯ β 4π₯ + 3 and group as 8π₯ 2 β 6π₯ + β4π₯ + 3 β’ Factor the GCF from each group: 2π₯ 4π₯ β 3 β 1 β 4π₯ β 3 = (4π₯ β 3)(2π₯ β 1) 16 2 Solving ππ₯ + ππ₯ + π = 0 by Factoring β’ To solve the quadratic equation 8π₯ 2 β 10π₯ + 3 = 0 (note that it must equal zero!), we use a property of numbers involving zero β’ Product Property of Zero: If ππ = 0, then π = 0 or π = 0 β’ That is, if you multiply two numbers and the result is zero, then one or the other (or both) of the numbers must be zero β’ We can use this to solve by factoring because factoring an expression makes the expression a multiplication β’ This is why the equation must be equal to zero!!! 17 2 Solving ππ₯ + ππ₯ + π = 0 by Factoring β’ Letβs use this property to solve 8π₯ 2 β 10π₯ + 3 = 0 β’ You already know that 8π₯ 2 β 10π₯ + 3 = 4π₯ β 3 2π₯ β 1 β’ So we rewrite the equation as 4π₯ β 3 2π₯ β 1 = 0 β’ Using the zero property ππ = 0 βΉ π = 0 or π = 0, the expression 4π₯ β 3 is like the π and the expression 2π₯ β 1 is like the π β’ At this point, we can split the equation into two equations β’ Either 4π₯ β 3 = 0 or 2π₯ β 1 = 0 (note that these are both linear equations) β’ Finally, solve each linear equation to get π₯ = 3 4 or π₯ = 1 2 18 2 Solving ππ₯ + ππ₯ + π = 0 by Factoring β’ Now check the results 3 8 4 2 3 β 10 +3= 4 9 30 8 β +3= 16 4 9 30 18 30 β12 β +3= β +3= + 3 = β3 + 3 = 0 2 4 4 4 4 19 2 Solving ππ₯ + ππ₯ + π = 0 by Factoring β’ Now check the results 1 8 2 2 1 β 10 +3= 2 1 8 β5+3= 4 2 β 5 + 3 = β3 + 3 = 0 20 2 Solving ππ₯ + ππ₯ + π = 0 by Factoring β’ Example: solve π₯ 2 β 2π₯ = 48 β’ The equation must equal zero; subtract 48 from both sides of the equation π₯ 2 β 2π₯ β 48 = 0 β’ Factor the left side of the equation π₯+6 π₯β8 =0 β’ Create two linear equations by setting each factor equal to zero (the product property of zero) π₯ + 6 = 0 or π₯ β 8 = 0 β’ Solve each linear equation π₯ = β6 or π₯ = 8 21 2 Solving ππ₯ + ππ₯ + π = 0 by Factoring β’ Example: solve π₯ 2 = 8π₯ β 16 β’ The equation must equal zero; subtract 8π₯ and add 16 π₯ 2 β 8π₯ + 16 = 0 β’ Factor the left side of the equation π₯β4 π₯β4 =0 β’ Create two linear equations by setting each factor equal to zero (the product property of zero) π₯ β 4 = 0 or π₯ β 4 = 0 β’ Solve each linear equation (note that this equation has only one solution) π₯=4 22 2 Solving ππ₯ + ππ₯ + π = 0 by Factoring β’ We can also use the square root property to solve this equation (after it has been factored) β’ Since (π₯ β 4)(π₯ β 4) is the same as π₯ β 4 2 , then the equation becomes π₯β4 2 =0 β’ By the square root property, π₯ β 4 = ±0 = 0 β’ Therefore, π₯ = 4 is the only solution β’ We will use the square root property in this way when you learn one more method for solving quadratic equations 23 2 Solving ππ₯ + ππ₯ + π = 0 by Factoring β’ Example: solve π₯ 2 β 2π₯ β 2 = 0 β’ β’ β’ β’ β’ β’ This is already equal to zero, so no change is needed in the equation There are no integers that multiply to give β2 and also add to give β2 This does not mean that this has no solution! It means that we cannot solve this by factoring The two solutions are π₯ = 1 + 2 and π₯ = 1 β 2 You will learn how to solve these by other methods (and this type of problem will not appear in your practice) 24 2 Solving ππ₯ + ππ₯ + π = 0 by Factoring β’ Example: solve 2π₯ 2 + 7π₯ + 3 = 0 β’ This is already equal to zero, so no change is needed in the equation β’ Multiply the lead coefficient by the constant to get 6: choose 6 and 1 since 6 β 1 = 6 and 6 + 1 = 7 β’ Rewrite the equation as 2π₯ 2 + 6π₯ + π₯ + 3 = 0 β’ Factor by grouping 2π₯ 2 + 6π₯ + π₯ + 3 = 2π₯ π₯ + 3 + 1 π₯ + 3 = (2π₯ + 1)(π₯ + 3) β’ The equation is now 2π₯ + 1 π₯ + 3 = 0 β’ Set each factor equal to zero and solve the two linear equations 2π₯ + 1 = 0 or π₯ + 3 = 0 1 β’ The solutions are π₯ = β or π₯ = β3 2 25 2 Solving ππ₯ + ππ₯ + π = 0 by Factoring β’ Example: solve β10π 2 = 11π β 6 β’ This is not equal to zero; it is easier to factor if the lead coefficient is not negative, so add 10π 2 to both sides 0 = 10π 2 + 11π β 6 or 10π 2 + 11π β 6 = 0 β’ Factor the expression: 10 β β6 = β60; choose 15 and β4 since 15 β β4 = β 60 and 15 + β4 = 11 β’ Factor by grouping 10π 2 + 15π β 4π β 6 = 0 10π 2 + 15π + β4π β 6 = 0 5π 2π + 3 β 2 2π + 3 = 0 2π + 3 5π β 2 = 0 26 2 Solving ππ₯ + ππ₯ + π = 0 by Factoring β’ Example: solve β10π 2 = 11π β 6 β’ Set each factor equal to zero and solve the linear equations 5π₯ β 2 = 0 or 2π₯ + 3 = 0 2 3 π₯ = or π₯ = β 5 2 27 2 Solving ππ₯ + ππ₯ = 0 β’ Note that a quadratic equation of the form ππ₯ 2 + ππ₯ = 0 is easy to solve because we will always be able to factor an π₯ β’ Example: solve 4π₯ 2 β 8π₯ = 0 β’ We can find the GCF and rewrite the equation as 4π₯ π₯ β 2 = 0 β’ Now we again apply the zero product property with π being 4π₯ and π as π₯ β 2 4π₯ = 0 or π₯ β 2 = 0 π₯ = 0 or π₯ = 2 28 Guided Practice Solve the following quadratic equations by factoring. a) 6π₯ 2 = 5π₯ b) π₯ 2 β 8π₯ + 15 = 0 c) 10π₯ β 3 = β8π₯ 2 d) 7π₯ 2 + 9π₯ = β2 e) π₯ 2 β 12π₯ + 36 = 0 f) 8π₯ 2 β 16π₯ = 0 29 Guided Practice Solve the following quadratic equations by factoring. a) 6π₯ 2 = 5π₯, π₯ = 0 or π₯ = 5 6 b) π₯ 2 β 8π₯ + 15 = 0, π₯ = 3 or π₯ = 5 c) 10π₯ β 3 = β8π₯ β3 1 ,π₯= or π₯ = 2 4 β2 π₯= or π₯ = β1 7 2 d) 7π₯ 2 + 9π₯ = β2, e) π₯ 2 β 12π₯ + 36 = 0, π₯ = 6 f) 8π₯ 2 β 16π₯ = 0, π₯ = 0 or π₯ = 2 30