Why isn't the definition of absolute value applied when squaring a radical containing a variable?Different...

Could moose/elk survive in the Amazon forest?

Combinable filters

Don’t seats that recline flat defeat the purpose of having seatbelts?

Check if a string is entirely made of the same substring

Contradiction proof for inequality of P and NP?

What's a good approximation to the zeros of the cosine integral?

What is Niska's accent?

Pre-plastic human skin alternative

TIKZ - changing one block into parallel multiple blocks

What is the origin of the term "plane of the spirit level"?

Why was Germany not as successful as other Europeans in establishing overseas colonies?

Apply MapThread to all but one variable

Multiple options vs single option UI

Does holding a wand and speaking its command word count as V/S/M spell components?

What term is being referred to with "reflected-sound-of-underground-spirits"?

A ​Note ​on ​N!

Do I have to worry about players making “bad” choices on level up?

Disappearing left side of `aligned`

How to limit Drive Letters Windows assigns to new removable USB drives

"The cow" OR "a cow" OR "cows" in this context

Can someone publish a story that happened to you?

How to reduce LED flash rate (frequency)

Why other Westeros houses don't use wildfire?

The Defining Moment



Why isn't the definition of absolute value applied when squaring a radical containing a variable?


Different ways of defining Absolute ValueAdding $2$ absolute values together: $|x+2| + |x-3| =5.$Why does $|x_1| = |x_2| implies x_1 = pm x_2$Why does the integral $intfrac{1}{x+i}dx$ not require the absolute value in the logarithm?The Definition of the Absolute ValueSpivak's Calculus 4th ed: Absolute value definition p. 11Solving Equations with Roots and Absolute ValuesAbsolute Value. Real AnalysisSolving Radical InequalitiesHow to get rid of the absolute value when solving a system of complex equations?













1












$begingroup$


I recently learned about the following definition of absolute value:




$|a| = sqrt{a^2}$




Then I came across a solution to a problem that had the following step:




$5 geq sqrt{5 - x}$




In order to proceed, we had to square both sides:




$5^2 geq (sqrt{5 - x})^2$




With the aforementioned definition of absolute value in mind, I wrote:




$25 geq |5 - x|$




But the actual solution turned out to be:




$25 geq 5 - x$




I don't understand why the absolute value definition wasn't applied here. Can anyone tell me why?






algebra-precalculus radicals absolute-value







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    It is applied. What you need to note is that the operation of squaring both sides of an inequality is not an equivalent transformation, like adding the same amount on both sides is. The inequality before squaring implies the inequality after squaring, but the reverse implication is not satisfied. So, when you square the new inequality has all solutions of the original, but it might have additional solutions. So, when you square you do apply the absolute value, since this is in the definition of $sqrt{ }$, but after solving the resulting inequality, you need to check for false solutions.
    $endgroup$
    – user647486
    1 hour ago


















1












$begingroup$


I recently learned about the following definition of absolute value:




$|a| = sqrt{a^2}$




Then I came across a solution to a problem that had the following step:




$5 geq sqrt{5 - x}$




In order to proceed, we had to square both sides:




$5^2 geq (sqrt{5 - x})^2$




With the aforementioned definition of absolute value in mind, I wrote:




$25 geq |5 - x|$




But the actual solution turned out to be:




$25 geq 5 - x$




I don't understand why the absolute value definition wasn't applied here. Can anyone tell me why?






algebra-precalculus radicals absolute-value







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    It is applied. What you need to note is that the operation of squaring both sides of an inequality is not an equivalent transformation, like adding the same amount on both sides is. The inequality before squaring implies the inequality after squaring, but the reverse implication is not satisfied. So, when you square the new inequality has all solutions of the original, but it might have additional solutions. So, when you square you do apply the absolute value, since this is in the definition of $sqrt{ }$, but after solving the resulting inequality, you need to check for false solutions.
    $endgroup$
    – user647486
    1 hour ago
















1












1








1





$begingroup$


I recently learned about the following definition of absolute value:




$|a| = sqrt{a^2}$




Then I came across a solution to a problem that had the following step:




$5 geq sqrt{5 - x}$




In order to proceed, we had to square both sides:




$5^2 geq (sqrt{5 - x})^2$




With the aforementioned definition of absolute value in mind, I wrote:




$25 geq |5 - x|$




But the actual solution turned out to be:




$25 geq 5 - x$




I don't understand why the absolute value definition wasn't applied here. Can anyone tell me why?






algebra-precalculus radicals absolute-value







share|cite|improve this question









$endgroup$




I recently learned about the following definition of absolute value:




$|a| = sqrt{a^2}$




Then I came across a solution to a problem that had the following step:




$5 geq sqrt{5 - x}$




In order to proceed, we had to square both sides:




$5^2 geq (sqrt{5 - x})^2$




With the aforementioned definition of absolute value in mind, I wrote:




$25 geq |5 - x|$




But the actual solution turned out to be:




$25 geq 5 - x$




I don't understand why the absolute value definition wasn't applied here. Can anyone tell me why?






algebra-precalculus radicals absolute-value




algebra-precalculus radicals absolute-value






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 1 hour ago









CalculemusCalculemus

392217




392217








  • 1




    $begingroup$
    It is applied. What you need to note is that the operation of squaring both sides of an inequality is not an equivalent transformation, like adding the same amount on both sides is. The inequality before squaring implies the inequality after squaring, but the reverse implication is not satisfied. So, when you square the new inequality has all solutions of the original, but it might have additional solutions. So, when you square you do apply the absolute value, since this is in the definition of $sqrt{ }$, but after solving the resulting inequality, you need to check for false solutions.
    $endgroup$
    – user647486
    1 hour ago
















  • 1




    $begingroup$
    It is applied. What you need to note is that the operation of squaring both sides of an inequality is not an equivalent transformation, like adding the same amount on both sides is. The inequality before squaring implies the inequality after squaring, but the reverse implication is not satisfied. So, when you square the new inequality has all solutions of the original, but it might have additional solutions. So, when you square you do apply the absolute value, since this is in the definition of $sqrt{ }$, but after solving the resulting inequality, you need to check for false solutions.
    $endgroup$
    – user647486
    1 hour ago










1




1




$begingroup$
It is applied. What you need to note is that the operation of squaring both sides of an inequality is not an equivalent transformation, like adding the same amount on both sides is. The inequality before squaring implies the inequality after squaring, but the reverse implication is not satisfied. So, when you square the new inequality has all solutions of the original, but it might have additional solutions. So, when you square you do apply the absolute value, since this is in the definition of $sqrt{ }$, but after solving the resulting inequality, you need to check for false solutions.
$endgroup$
– user647486
1 hour ago






$begingroup$
It is applied. What you need to note is that the operation of squaring both sides of an inequality is not an equivalent transformation, like adding the same amount on both sides is. The inequality before squaring implies the inequality after squaring, but the reverse implication is not satisfied. So, when you square the new inequality has all solutions of the original, but it might have additional solutions. So, when you square you do apply the absolute value, since this is in the definition of $sqrt{ }$, but after solving the resulting inequality, you need to check for false solutions.
$endgroup$
– user647486
1 hour ago












4 Answers
4






active

oldest

votes


















3












$begingroup$

$$left(sqrt aright)^2nesqrt{a^2}.$$



Try with $a=-1$.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Indeed, $sqrt{a^2}=lvert arvert$. But $sqrt a^2=a$ (assuming that $ageqslant0$), not $lvert arvert$.






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      The inequality has the domain $$5geq x$$, so we can square it and we get $$25geq 5-x$$






      share|cite|improve this answer









      $endgroup$





















        1












        $begingroup$

        From the fact that you can take $sqrt {5-x}$ you know that $5-x ge 0$ so you don't need the absolute value signs.






        share|cite|improve this answer









        $endgroup$









        • 1




          $begingroup$
          You don't really "know that" $5-x ge 0$, rather you "must have" $5-x ge 0$; in either case, it follows that $|5-x| = 5-x$. So $25 ge |5-x|$ becomes $25 ge 5-x$.
          $endgroup$
          – steven gregory
          1 hour ago












        Your Answer








        StackExchange.ready(function() {
        var channelOptions = {
        tags: "".split(" "),
        id: "69"
        };
        initTagRenderer("".split(" "), "".split(" "), channelOptions);

        StackExchange.using("externalEditor", function() {
        // Have to fire editor after snippets, if snippets enabled
        if (StackExchange.settings.snippets.snippetsEnabled) {
        StackExchange.using("snippets", function() {
        createEditor();
        });
        }
        else {
        createEditor();
        }
        });

        function createEditor() {
        StackExchange.prepareEditor({
        heartbeatType: 'answer',
        autoActivateHeartbeat: false,
        convertImagesToLinks: true,
        noModals: true,
        showLowRepImageUploadWarning: true,
        reputationToPostImages: 10,
        bindNavPrevention: true,
        postfix: "",
        imageUploader: {
        brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
        contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
        allowUrls: true
        },
        noCode: true, onDemand: true,
        discardSelector: ".discard-answer"
        ,immediatelyShowMarkdownHelp:true
        });


        }
        });














        draft saved

        draft discarded


















        StackExchange.ready(
        function () {
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3204507%2fwhy-isnt-the-definition-of-absolute-value-applied-when-squaring-a-radical-conta%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        $$left(sqrt aright)^2nesqrt{a^2}.$$



        Try with $a=-1$.






        share|cite|improve this answer









        $endgroup$


















          3












          $begingroup$

          $$left(sqrt aright)^2nesqrt{a^2}.$$



          Try with $a=-1$.






          share|cite|improve this answer









          $endgroup$
















            3












            3








            3





            $begingroup$

            $$left(sqrt aright)^2nesqrt{a^2}.$$



            Try with $a=-1$.






            share|cite|improve this answer









            $endgroup$



            $$left(sqrt aright)^2nesqrt{a^2}.$$



            Try with $a=-1$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 1 hour ago









            Yves DaoustYves Daoust

            134k676232




            134k676232























                2












                $begingroup$

                Indeed, $sqrt{a^2}=lvert arvert$. But $sqrt a^2=a$ (assuming that $ageqslant0$), not $lvert arvert$.






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  Indeed, $sqrt{a^2}=lvert arvert$. But $sqrt a^2=a$ (assuming that $ageqslant0$), not $lvert arvert$.






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    Indeed, $sqrt{a^2}=lvert arvert$. But $sqrt a^2=a$ (assuming that $ageqslant0$), not $lvert arvert$.






                    share|cite|improve this answer









                    $endgroup$



                    Indeed, $sqrt{a^2}=lvert arvert$. But $sqrt a^2=a$ (assuming that $ageqslant0$), not $lvert arvert$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 1 hour ago









                    José Carlos SantosJosé Carlos Santos

                    178k24139251




                    178k24139251























                        1












                        $begingroup$

                        The inequality has the domain $$5geq x$$, so we can square it and we get $$25geq 5-x$$






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          The inequality has the domain $$5geq x$$, so we can square it and we get $$25geq 5-x$$






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            The inequality has the domain $$5geq x$$, so we can square it and we get $$25geq 5-x$$






                            share|cite|improve this answer









                            $endgroup$



                            The inequality has the domain $$5geq x$$, so we can square it and we get $$25geq 5-x$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 1 hour ago









                            Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                            79.9k42867




                            79.9k42867























                                1












                                $begingroup$

                                From the fact that you can take $sqrt {5-x}$ you know that $5-x ge 0$ so you don't need the absolute value signs.






                                share|cite|improve this answer









                                $endgroup$









                                • 1




                                  $begingroup$
                                  You don't really "know that" $5-x ge 0$, rather you "must have" $5-x ge 0$; in either case, it follows that $|5-x| = 5-x$. So $25 ge |5-x|$ becomes $25 ge 5-x$.
                                  $endgroup$
                                  – steven gregory
                                  1 hour ago
















                                1












                                $begingroup$

                                From the fact that you can take $sqrt {5-x}$ you know that $5-x ge 0$ so you don't need the absolute value signs.






                                share|cite|improve this answer









                                $endgroup$









                                • 1




                                  $begingroup$
                                  You don't really "know that" $5-x ge 0$, rather you "must have" $5-x ge 0$; in either case, it follows that $|5-x| = 5-x$. So $25 ge |5-x|$ becomes $25 ge 5-x$.
                                  $endgroup$
                                  – steven gregory
                                  1 hour ago














                                1












                                1








                                1





                                $begingroup$

                                From the fact that you can take $sqrt {5-x}$ you know that $5-x ge 0$ so you don't need the absolute value signs.






                                share|cite|improve this answer









                                $endgroup$



                                From the fact that you can take $sqrt {5-x}$ you know that $5-x ge 0$ so you don't need the absolute value signs.







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered 1 hour ago









                                Ross MillikanRoss Millikan

                                302k24201375




                                302k24201375








                                • 1




                                  $begingroup$
                                  You don't really "know that" $5-x ge 0$, rather you "must have" $5-x ge 0$; in either case, it follows that $|5-x| = 5-x$. So $25 ge |5-x|$ becomes $25 ge 5-x$.
                                  $endgroup$
                                  – steven gregory
                                  1 hour ago














                                • 1




                                  $begingroup$
                                  You don't really "know that" $5-x ge 0$, rather you "must have" $5-x ge 0$; in either case, it follows that $|5-x| = 5-x$. So $25 ge |5-x|$ becomes $25 ge 5-x$.
                                  $endgroup$
                                  – steven gregory
                                  1 hour ago








                                1




                                1




                                $begingroup$
                                You don't really "know that" $5-x ge 0$, rather you "must have" $5-x ge 0$; in either case, it follows that $|5-x| = 5-x$. So $25 ge |5-x|$ becomes $25 ge 5-x$.
                                $endgroup$
                                – steven gregory
                                1 hour ago




                                $begingroup$
                                You don't really "know that" $5-x ge 0$, rather you "must have" $5-x ge 0$; in either case, it follows that $|5-x| = 5-x$. So $25 ge |5-x|$ becomes $25 ge 5-x$.
                                $endgroup$
                                – steven gregory
                                1 hour ago


















                                draft saved

                                draft discarded




















































                                Thanks for contributing an answer to Mathematics Stack Exchange!


                                • Please be sure to answer the question. Provide details and share your research!

                                But avoid …



                                • Asking for help, clarification, or responding to other answers.

                                • Making statements based on opinion; back them up with references or personal experience.


                                Use MathJax to format equations. MathJax reference.


                                To learn more, see our tips on writing great answers.




                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function () {
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3204507%2fwhy-isnt-the-definition-of-absolute-value-applied-when-squaring-a-radical-conta%23new-answer', 'question_page');
                                }
                                );

                                Post as a guest















                                Required, but never shown





















































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown

































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown







                                -

                                Popular posts from this blog

                                Щит и меч (фильм) Содержание Названия серий | Сюжет |...

                                Image
                                Фильмы по алфавитуФильмы 1968 годаФильмы СССР 1968 годаЧёрно-белые фильмы СССРФильмы студии «Мосфильм»Лидеры советского кинопрокатаФильмы-драмы СССРФильмы о Второй мировой войнеФильмы о шпионахЭкранизации произведений Вадима КожевниковаФильмы Владимира БасоваФильмы о фашизме 1968 годуодноимённому романуВадима КожевниковаВладимиром БасовымСтанислав ЛюбшинГерманиюАбверегауптштурмфюрераССБерлинслужбу СДSchwarzbraun ist die HaselnussXVIII векаЭрикиХермса НиляГитлерюгендаАбвераАх, мой милый АвгустинXVII векаHorst WesselDeutschland über alles У этого термина существуют и другие значения, см. Щит и меч. Щит и меч Жанр драма военный фильм Режиссёр Владимир Басов Автор сценария Владимир Басов Вадим Кожевников В главных ролях Станислав Любшин Валентина Титова Олег Янковский Георгий Мартынюк Владимир Басов Оператор Тимофей Лебешев Композитор Вениамин Баснер Кинокомпания Киностудия «Мосфильм». Творческое объединение писателей и киноработников
                                Read more

                                Венесуэла на летних Олимпийских играх 2000 Содержание Состав...

                                Image
                                БангладешБахрейнБутанБрунейВьетнамГонконгИндияИндонезияИорданияИракИранЙеменКазахстанКамбоджаКатарКитайКиргизияКНДРКувейтЛаосЛиванМалайзияМальдивыМонголияМьянмаНепалОАЭОман ПакистанГосударство ПалестинаСаудовская АравияСингапурСирияТаджикистанТаиландТайваньТуркменияУзбекистанФилиппиныШри-ЛанкаЮжная КореяЯпонияИндивидуальные олимпийские спортсменыАлжирАнголаБенинБотсванаБуркина-ФасоБурундиГабонГамбияГанаГвинеяГвинея-БисауДемократическая Республика КонгоДжибутиЕгипетЗамбияЗимбабвеКабо-ВердеКамерунКенияКоморские островаКот-д’ИвуарЛесотоЛиберияЛивияМаврикийМавританияМадагаскарМалавиМалиМароккоМозамбикНамибияНигерНигерияРеспублика КонгоРуандаСан-Томе и ПринсипиСвазилендСейшелыСенегалСомалиСуданСьерра-ЛеонеТанзанияТогоТунисУгандаЦентральноафриканская РеспубликаЧадЭкваториальная ГвинеяЭритреяЭфиопияЮАРАвстралияАмериканское СамоаВануатуГуамМикронезияНауруНовая ЗеландияОстрова КукаПалауПапуа — Новая ГвинеяСамоаСоломоновы ОстроваТонгаФиджи Венесуэла на Олимпийских играх2000 год в ВенесуэлеСтран
                                Read more

                                Meter-Bus Содержание Параметры шины | Стандартизация |...

                                Image
                                RS-232IrDAHARTмодемтоковая петляИРПСMIDIIO-LinkRS-422RS-423RS-485LINM-BusKansas City StandardКомпакт-кассетагрампластинкаPPPSLIPIrDAHARTISO 7816MIDIIO-LinkModbusLINDMX-512P-NETProfibusIEC 608708250 UART16550 UART Сетевые протоколыПоследовательные шины физического уровняполевой шиныасинхронного интерфейсаOpen Metering SystemAESздесь У этого термина существуют и другие значения, см. M-bus (значения). M-Bus (Meter-Bus) — стандарт физического уровня для полевой шины на основе асинхронного интерфейса. Также под этим названием понимают коммуникационный протокол, используемый для связи устройств по этой шине. M-bus преимущественно применяется для приборов учета электрической энергии (электросчётчики), тепловой энергии (теплосчётчики), расходомеров воды и газа. Данные передаются на компьютерную станцию (сервер) напрямую или через концентраторы шины M-Bus, а также усилители-повторители сигнала. Содержание 1 Параметры шины 2 Стандартизация 2.1 Open Metering System
                                Read more