Package Summary. Square Root Equations Absolute Value Equations. Making Math Possible 1 of 9 c Sa diyya Hendrickson

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1 Solving Equations Preared by: Sa diyya Hendrickson Name: Date: Package Summary Square Root Equations Absolute Value Equations Making Math Possible 1 of 9 c Sa diyya Hendrickson

2 A square root equation is an equation with olynomial terms along with at least one square root exression (with a olynomial under the radical). Consider the following examle: x +=x Recall that the ultimate goal when solving an equation is to isolate the variable. Answer: One of our x s is being held hostage by the square root! So, even before we begin, we know that at some oint we ll need to do the oosite of taking a square root (i.e. squaring), in order to free x! 1 Isolate the radical before squaring both sides. Why? Consider the equation a ± b = c. (F) e.g. If a =, b = x + and c =x, we can create the equation x +=x, which was given as the revious examle. Now, if we square both sides of the equation given in (F), we have: a ± b = c ) (a ± b) =(c) ) (a ± b)(a ± b)=c ) a ± a b ± a b +(± b) = c ) a ± a b + b = c Unfortunately, the distributive roerty guarantees that the radical will reaear in the exansion, articularly in the middle term. On the other hand, if we isolate the radical first, we have: ± b = c a Making Math Possible of 9 c Sa diyya Hendrickson

3 Squaring both sides of this equation gives: ± b = c a ) (± b) =(c a) ) b =(c a) Now it s clear that we ve achieved our goal: undoing the square root! Moral of the Story Isolate the radical first! If we square both sides of an equation, we are creating a new equation that may introduce additional, invalid (extraneous) solutions. Therefore, we must always check our answers at the end to make sure that L.S. = R.S. for the original equation. Why does this haen? Suose we are given some equation: a = b (E 1 ) If we square both sides of E 1, we create a new equation: a = b (E ) Answer: FromE we have: a = b () a = b. () a = b In words, E is satisfied whenever a and b are the same distance from zero on the real line. This results in exactly two outcomes: (1) a and b are the same. i.e. a = b () a and b are oosites. i.e. a = b Notice that these outcomes result in our need to solve two equations, one of which is E 1 and another that is not. So, solving E will always roduce the solutions to E 1 while otentially introducing additional solutions that are invalid. Moral of the Story Although squaring both sides hels us to free a variable from a square root, we must ay the fee of checking our solutions at the end to ensure that we re only taking valid solutions for the original equation! Making Math Possible 3 of 9 c Sa diyya Hendrickson

4 The key observations reviously discussed uncover some strategies that may be helful as we solve square root equations. Consider the following worked examles: Exercise SR1: Solve the equation x +=x. S1 Isolate the term involving a radical! Note: If there is more than one, isolate one radical, erform S and square both sides. Then, exand, collect and reeat this sequence for the others. x +=x () x += x S Square both sides to undo the square root oeration, and write a reminder at the bottom of your age to check solutions. Squaring both sides, we have: ( x +) = ( x) ) x + =( x) S3 Identify the new equation and roceed with all relevant strategies. This is a quadratic equation! So, first stategy: get a zero on one side! x +=( x) ) 0=( x) x ) ( x)( x) x =0 ) 4 4x 4x +4x x =0 collect ) 4x 9x +=0 and D =( 9) 4(4)() = 49 = 7 factor ) (x )(4x 1) = 0 ZPP ) x =0 or 4x 1=0 ) x = or x = 1 4 S4 Check that each value satisfies the original equation. i.e. Substitute each value and verify whether L.S. = R.S. For x =: L.S. = () + R.S. = () = 4 = 4 = =0 Because 0 6= 4, L.S.6= R.S. Thus, x = isinvalid/extraneous. Making Math Possible 4 of 9 c Sa diyya Hendrickson

5 Exercise SR1 (Continued) For x = 1 4 : L.S. = q LCD = = = R.S. = 1 4 q 1+()(4) 4 = LCD = () 3 = 4 3 = 1 ) L.S. = R.S. Therefore, x = 1 4 is the only solution. Exercise SR: Solve the equation x +5 3x =1 Solution: Notice that we have two terms involving a radical. So, we ll have to carry out S1 followed by S twice! S1 Isolating the first radical, we have: x +5=1+ 3x. S Squaring both sides and writing our check reminder, we have: ( x +5) = (1+ 3x ) ) x +5 = (1+ 3x )(1 + 3x ) ) x +5 = 1+ 3x + 3x {z } +( 3x ) collect ) x +5 = 1+ 3x +3x S1 Isolating the next radical term and collecting gives: 3x =6 x S Squaring both sides again, we have: ( 3x ) = (6 x) ) 4(3x ) = (6 x) a quadratic equation! Making Math Possible 5 of 9 c Sa diyya Hendrickson

6 Exercise SR (Continued) S3 Identifying and solving the quadratic equation, we have: 4(3x ) = (6 x) ) 0=(6 x) 4(3x ) S4 Checking our results, we have: ) (6 x)(6 x) 4(3x ) = 0 ) 36 6x 6x + x 1x +8=0 collect ) x 4x +44=0 and D =( 4) 4(1)(44) > 0 factor ) (x )(x ) = 0 ZPP ) x =0 or x = 0 ) x = or x = For x =: L.S. = () + 5 3() R.S. = 1 = = 9 4 = 3 = 1 L.S. = R.S. So, x = isavalidsolution! For x =: L.S. = () + 5 3() R.S. = 1 = = = 7 8 = 1 Because 1 6= 1, L.S.6= R.S. Thus, x = is invalid/extraneous. Therefore, x = is the only solution. Making Math Possible 6 of 9 c Sa diyya Hendrickson

7 Absolute Value Equations Consider the following examles of absolute value equations: (a) 3x 5 3=1 (b) x 1 =x +4 (c) 3 x = x 5 Answer: x is in jail, behind absolute value bars. So, to free x we must use the definition of absolute value! Note that there are di erent tyes of absolute value equations. Below we will discuss two, both of which have only one term involving absolute value. Tye I One absolute value term and all other terms are constants. Exercise AV1: Solve the equation 3x 5 3 = 1 S1 Isolate the absolute value term by moving all constants/numbers to the other side. i.e. Create an equation of the form: a = b. 3x 5 3=1 () 3x 5 =4 S Use the definition of absolute value to roduce the relevant cases. In general, for some nonnegative constant b, a = b means that a is a distance of b units from zero. So, either a = b (Case 1) or a = b (Case ). We have 3x 5 =4,whichmeansthat a =3x 5 is a distance of 4 units from zero! This gives the following two cases: Case 1 : 3x 5=4 Case : 3x 5= 4 Making Math Possible 7 of 9 c Sa diyya Hendrickson

8 Absolute Value Equations Exercise AV1 (Continued) S3 Solve the equation under each case. Case 1 Case 3x 5 = 4 3x 5 = x 3 = 9 3 3x 3 = 1 3 ) x =3 or x = 1 3 Tye II One absolute value term along with at least one non-constant term (i.e. a term involving the variable to some non-zero ower). Consider the following examle: x 1 4 = x. Notice that if we isolate the absolute value term, we have the equation x 1 =x The distance from 0 is reresented by x + 4, which changes as x changes. Because distance is a nonnegative quantity, x + 4 is restricted. Consequently, there are restrictions on x. These di erences require us to roceed with caution. A oular and safe aroach is to take cases as dictated by the algebraic definition of absolute value. While doing so, we must make note of the restrictions and check all solutions! Recall that: 8 < a if a 0 (Case 1) a = : a if a<0 (Case ) Exercise AV: Solve the equation x 1 = x + 4. S1 Use the algebraic definition of absolute value to determine the relevant cases and restrictions for the equation. 8 < x 1 if x 1 0 (Case 1) x 1 = : (x 1) if x 1 < 0 (Case ) Making Math Possible 8 of 9 c Sa diyya Hendrickson

9 Absolute Value Equations Exercise AV (Continued) S Solve the equation under each case. Be sure to identify and state the restrictions before solving! Case 1: x 1 0 Case : x 1 < 0 ) x 1 ) x <1 ) x 1 ) x< 1 Note: A solution is valid under each case only if it lies in the intervals stated above! Here, x 1 def n = x 1 Here, x 1 def n = (x 1) ) the equation becomes: ) the equation becomes: x 1 = x +4 (x 1) = x +4 x x ) x +1 = x +4 1 = 4 x = x This is a contradiction! So, in this interval, there is no solution! 4x +1 = 4 4x 4 1 = 1 = 3 4 ) x = 3 4 < 1 Therefore, x = 3 4 is the only solution for this equation. Making Math Possible 9 of 9 c Sa diyya Hendrickson