Solving Equations Prepared by: Sa diyya Hendrickson Name: Date: Package Summary Linear Equations Quadratic Equations Rational Equations Making Math Possible 1 of 10 c Sa diyya Hendrickson
Linear Equations I. Assigned Reading If you need to refresh your memory on linear equations, please read the Algebra Lesson (pages 10-13), which can be found on the Extra Help page of the course website: www.pumpuoft.yolasite.com/extra-help.php II. Summarizing our Strategies: S1 Identify which side of the equation x is on. If there is more than one x, decide which side you want to move all of them to. S2 Expand (if necessary) and do the opposite to rearrange the equation so that x is on one side (by itself) and numbers are on the other side. S3 Sneak up on x! Theultimategoalistoisolatex, andthebestapproachisto get everything out of x s way, starting from the numbers furthest away! Solve the following equation: 2(x 3) x = x +1 2 4 Making Math Possible 2 of 10 c Sa diyya Hendrickson
Quadratic Equations Below are three popular quadratic forms: 1. Standard Form: a x 2 + b x + c where a, b, c are constants with a 6= 0. e.g. 2x 2 x 5; a =2,b= 1andc = 5 2. Factored Form: d (m x + s)(n x + t) where d, m, n, s, t are constants such that d 6= 0, m 6= 0 and n 6= 0. e.g. 3(x 5)(5x +2); d = 3, m=1, s = 5,n= 5 and t =2 3. Vertex Form: a (x h) 2 + k where a, h, k are constants with a 6= 0. e.g. 7 (x +1) 2 4=7(x ( 1)) 2 4; a =7,h= 1, and k = 4 Answer: One side of the equation can be expressed in a quadratic form, while the other side has a linear or quadratic form. e.g. x(x 4) = 5 2x ) Notice that after expansion, every term will involve a nonnegative power of x that is less than or equal to 2. (i.e. each term will involve x 0 =1,x 1 = x or x 2 ) Exercise: Which of the following is a quadratic equation? a x 2 +1=5 b (1 x 2 )(x +2)=3x 1 c x(5x) = x 3 4+2x 2 d 3(x 1) 2 =7 x e x +6=4(x 7) + 1 Answer: It depends! The possibilities for the number of real solutions are: no solution, one solution or two solutions. Let s take a look at the quadratic formula to get a good understanding of when and why each case occurs. Making Math Possible 3 of 10 c Sa diyya Hendrickson
Quadratic Equations Suppose we have a quadratic in standard form, as follows: a x 2 + b x + c. Now, consider the related equation of the form: a x 2 + b x + c =0 (1) The quadratic formula gives that the general solutions (if they exist) are: x ± = b ± p b 2 4ac 2a ) Let m = x + = b + p b 2 4ac 2a and n = x = b p b 2 4ac 2a The expression under the radical, namely D = b 2 4ac, iscalledthediscriminant and it is what determines the number of solutions/zeros/roots for the quadratic equation. Note: Thenumberofsolutionsalsoindicatesifthequadraticintheequationisreducible (i.e. can be factored). If the quadratic has no solutions/zeros/roots, we conclude that it cannot be factored in the reals and is called irreducible. Cases p D = p b 2 4ac Solutions #of solutions factored form D<0 undefined N/A 0 irreducible D =0 D>0 p p D = 0=0 m = n = b±(0) = b 1 (x m) 2 2a 2a p D>0 m and n 2 (x m)(x n) Remark: Thequadraticformulaiscomputationallyheavy,soitshouldbeusedasalast resort when factoring! Try factoring x 2 +x 6 using both trial and error and the formula. How many solutions does the equation x 2 + 1 = 0 have? Solution: First note that the quadratic equation is in the appropriate form described in (1). Furthermore, we can see that a =1,b =0 and c =1. Therefore,thediscriminant is given by: D =(0) 2 4(1)(1) = 4 < 0 The discriminant is negative. Thus,thisequationhasno real solutions. Making Math Possible 4 of 10 c Sa diyya Hendrickson
Quadratic Equations Let s recall the three popular quadratic forms once again: 1. Standard Form : a x 2 + b x + c 2. Factored Form : d (m x + s)(n x + t) 3. Vertex Form : a (x h) 2 + k For Vertex Form a (x h) 2 + k = ` where a, h, k and ` are constants. If you are given an equation with a quadratic in vertex form, notice that x is in one place, as we require for linear equations! So, we will simply isolate x by doing the opposite! Note: Every quadratic can be expressed in vertex form by completing the square (See handout). However, if you are given an equation with a quadratic that is not already in vertex form, it is not practical to complete the square in order to isolate x. Instead, proceed with the strategies beginning on page 8! Exercise Q1: Solve the quadratic equation: 3 (x 1) 2 4=2 S1 Isolate the square term!. ) 3(x 1)2 3 = 6 3 ) (x 1) 2 =2 S2 Do the opposite to undo the squaring operation. i.e. Take the square root of both sides and simplify! S3 Isolate x! (x 1) 2 =2 Sq.Root ) p (x 1) 2 = p 2 P.O.E ) x 1 = p 2 Def n ) x 1=± p 2 x 1 +1 = ± p 2 +1 ) x =1± p 2 Making Math Possible 5 of 10 c Sa diyya Hendrickson
Quadratic Equations Consider the following quadratic equation: 3 + x 2 =2x(x 1) Unlike linear equations and quadratics in vertex form, isolating x isn t as straightforward. Even if we move the terms involving x to one side of the equation, we won t be able to combine the square terms with the linear terms, since they aren t like terms. Remember that it is not practical to complete the square to achieve vertex form. Toget abitofguidanceonthebestnextstep,let srecalltwoofourpopularquadraticforms: 1. Standard Form : a x 2 + b x + c 2. Factored Form : d (m x + s)(n x + t) It s quite clear that if b 6= 0,standard form isn t very useful because we can t combine ax 2 with bx to get x in one place. So, we consider the factored and vertex forms. For Factored Form Notice that our factored form is a product of the terms A = d, B = mx + s and C = nx + t. Furthermore, if there is a way to create seperate linear equations involving A, B and C, thenwe llbeabletoisolate x very easily! Consider the Zero-Product Property (ZPP),whichtellsusthat if we have a product of numbers that equals 0 (for example, A B C =0),thenat least one of the numbers must be 0! In other words, A =0,B =0 or C =0,givingus3seperate equations! So, in order to use ZPP as a strategy, we need an equation with: 1 a zero on one side! 2 a product/factored form on the other side! This uncovers a few strategies for solving some quadratic equations! Making Math Possible 6 of 10 c Sa diyya Hendrickson
Quadratic Equations Exercise Q2: Solve the quadratic equation: 3 + x 2 =2x(x 1) S1 Get a zero on one side by doing the opposite!. Itmaybeworthwhileto move terms so that your x 2 term remains positive after collecting like terms. 3+x 2 =2x(x 1) () 0 =2x(x 1) 3 x 2 S2 Collect like terms! Note: sometimesyoumayneedtoexpand first. 0=2x(x 1) 3 x 2 DP ) 2x 2 2x 3 x 2 =0 collect ) 2x 2 x 2 2x 3=0 collect ) x 2 2x 3=0 S3 Check the discriminant D before attempting to factor! Sometimesyou may find that there are no solutions! a =1,b= 2 and c = 3 ) D =( 2) 2 4(1)( 3) = 16 > 0 S4 If D 0, factor by trial and error to make the quadratic look like a product (so that you can use ZPP), or use the quadratic formula. Note: IfD = b 2 4ac is a perfect square, trialanderrorwillalwayswork! Ifnot, then the roots are irrational and you ll need to use the quadratic formula. From S3, we know that D =16=4 2 (a perfect square). So, we will attempt to factor by trial and error (using the All-In-One Factoring Method). ) (x +1)(x 3) =0 {z } A B S5 If you factored, use ZPP to seperate the linear equations! ZPP ) x +1=0 or x 3=0 Making Math Possible 7 of 10 c Sa diyya Hendrickson
Rational Equations S6 Isolate x! Rational equations are equations involving rational expressions (i.e. fractions with at least one variable in a denominator). e.g. 2 x 1 =3+x +4 x Exercise R1: Solve the equation x = 9 2x x+3 x 2 +x 6 Below, we will discuss two approaches. For each one, we will uncover useful strategies! Approach 1 Use the fact that given any fraction a,if a =0 ) a =0 b b and b 6= 0. i.e. a numerator equalling zero and a non-zero denominator! S1 Get a zero on one side by doing the opposite. x x +3 = 9 2x x 2 + x 6 () x x +3 9 2x x 2 + x 6 = 0 S2 Get an LCD for the rational expression so that we can transform the sum/di erence into one fraction. D 1 = x +3 and D 2 = x 2 + x 6=(x +3)(x 2) ) LCD = (x +3)(x 2) Making the appropriate adjustments, we have: x x +3 9 2x x 2 + x 6 =0 LCD ) x (x 2) (x +3)(x 2) 9 2x (x +3)(x 2) =0 LCD ) x (x 2) (9 2x) (x +3)(x 2) =0 Making Math Possible 8 of 10 c Sa diyya Hendrickson
Rational Equations S3 Because we have a fraction equal to zero: 1 State restrictions (i.e. Omit values for which DENOM = 0) and 2 Set NUM = 0 and solve the new equation. 1 Restrictions: DENOM 6= 0 ) (x +3)(x 2) 6= 0 ZPP ) x +36= 0 and x 2 6= 0 ) x 6= 3, 2 2 NUM = 0 ) x (x 2) (9 2x) =0 DP ) x 2 2x 9+2x =0 collect ) x 2 9=0 factor ) (x +3)(x 3) = 0 ZPP ) x +3=0 or x 3=0 ) x = 3 or x =3 S4 Check that the solutions don t include any restricted values. Notice that x = 3 is not valid, since it appeared under the restrictions. Therefore, x =3 is the only solution. Approach 2 Use the fact that given any two fractions a and c,if a = c ) b d b d ad = bc (a.k.a. cross multiplication ). We are clearing the denominators by multiplying each side of the equation by a common denominator. The strategies for this approach are given below. S1 Determine the LCD among the terms on each side of the equation. Then, transform each side of the equation into one fraction with the required LCDs. In other words, get the form a = c. b d S2 State the restrictions for the fraction on each side and clear the denominators by cross multiplying. S3 Solve the new equation. S4 Check that the solutions do not include any restricted values. Making Math Possible 9 of 10 c Sa diyya Hendrickson
Rational Equations Exercise: Try to solve the previous exercise using the strategies for Approach 2. Then, determine which approach you like better! Note: Toseeaworked example using a similar approach, see the Precalculus textbook by Stewart, page 53, Example 10. Making Math Possible 10 of 10 c Sa diyya Hendrickson