Simplifying expressions with radical exponents












0












$begingroup$


This is the expression that needs to be simplified:



$frac{(9-x^2)^{1/2} + x^2(9-x^2)^{-1/2}}{9-x^2}$



My first step is to move the negative exponent to the denominator:



$frac{(9-x^2)^{1/2} + x^2}{(9-x^2)(9-x^2)^{1/2}}$



Then multiply the denominator and add the exponents:



$frac{(9-x^2)^{1/2} + x^2}{(9-x^2)^{3/2}}$



I don't know the next steps to reach the given answer of:



$frac{9}{(9-x^2)^{3/2}}$






algebra-precalculus radicals







share|cite|improve this question









$endgroup$












  • $begingroup$
    Your first step is not valid. That can only be done with factors of the whole numerator, not just factors of a term in the numerator.
    $endgroup$
    – MPW
    Dec 29 '18 at 16:23










  • $begingroup$
    $frac{9}{left(9-x^2right)^{3/2}}$
    $endgroup$
    – David G. Stork
    Dec 29 '18 at 16:53
















0












$begingroup$


This is the expression that needs to be simplified:



$frac{(9-x^2)^{1/2} + x^2(9-x^2)^{-1/2}}{9-x^2}$



My first step is to move the negative exponent to the denominator:



$frac{(9-x^2)^{1/2} + x^2}{(9-x^2)(9-x^2)^{1/2}}$



Then multiply the denominator and add the exponents:



$frac{(9-x^2)^{1/2} + x^2}{(9-x^2)^{3/2}}$



I don't know the next steps to reach the given answer of:



$frac{9}{(9-x^2)^{3/2}}$






algebra-precalculus radicals







share|cite|improve this question









$endgroup$












  • $begingroup$
    Your first step is not valid. That can only be done with factors of the whole numerator, not just factors of a term in the numerator.
    $endgroup$
    – MPW
    Dec 29 '18 at 16:23










  • $begingroup$
    $frac{9}{left(9-x^2right)^{3/2}}$
    $endgroup$
    – David G. Stork
    Dec 29 '18 at 16:53














0












0








0





$begingroup$


This is the expression that needs to be simplified:



$frac{(9-x^2)^{1/2} + x^2(9-x^2)^{-1/2}}{9-x^2}$



My first step is to move the negative exponent to the denominator:



$frac{(9-x^2)^{1/2} + x^2}{(9-x^2)(9-x^2)^{1/2}}$



Then multiply the denominator and add the exponents:



$frac{(9-x^2)^{1/2} + x^2}{(9-x^2)^{3/2}}$



I don't know the next steps to reach the given answer of:



$frac{9}{(9-x^2)^{3/2}}$






algebra-precalculus radicals







share|cite|improve this question









$endgroup$




This is the expression that needs to be simplified:



$frac{(9-x^2)^{1/2} + x^2(9-x^2)^{-1/2}}{9-x^2}$



My first step is to move the negative exponent to the denominator:



$frac{(9-x^2)^{1/2} + x^2}{(9-x^2)(9-x^2)^{1/2}}$



Then multiply the denominator and add the exponents:



$frac{(9-x^2)^{1/2} + x^2}{(9-x^2)^{3/2}}$



I don't know the next steps to reach the given answer of:



$frac{9}{(9-x^2)^{3/2}}$






algebra-precalculus radicals




algebra-precalculus radicals






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 29 '18 at 16:20









AntoninusAntoninus

33




33












  • $begingroup$
    Your first step is not valid. That can only be done with factors of the whole numerator, not just factors of a term in the numerator.
    $endgroup$
    – MPW
    Dec 29 '18 at 16:23










  • $begingroup$
    $frac{9}{left(9-x^2right)^{3/2}}$
    $endgroup$
    – David G. Stork
    Dec 29 '18 at 16:53


















  • $begingroup$
    Your first step is not valid. That can only be done with factors of the whole numerator, not just factors of a term in the numerator.
    $endgroup$
    – MPW
    Dec 29 '18 at 16:23










  • $begingroup$
    $frac{9}{left(9-x^2right)^{3/2}}$
    $endgroup$
    – David G. Stork
    Dec 29 '18 at 16:53
















$begingroup$
Your first step is not valid. That can only be done with factors of the whole numerator, not just factors of a term in the numerator.
$endgroup$
– MPW
Dec 29 '18 at 16:23




$begingroup$
Your first step is not valid. That can only be done with factors of the whole numerator, not just factors of a term in the numerator.
$endgroup$
– MPW
Dec 29 '18 at 16:23












$begingroup$
$frac{9}{left(9-x^2right)^{3/2}}$
$endgroup$
– David G. Stork
Dec 29 '18 at 16:53




$begingroup$
$frac{9}{left(9-x^2right)^{3/2}}$
$endgroup$
– David G. Stork
Dec 29 '18 at 16:53










3 Answers
3






active

oldest

votes


















0












$begingroup$

Hint: Write $$frac{(9-x^2)^{1/2}}{9-x^2}+frac{x^2(9-x^2)^{-1/2}}{9-x^2}$$ and this is
$$frac{1}{sqrt{9-x^2}}+frac{x^2}{(9-x^2)^{3/2}}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for the feedback. I understand the right side of this, but I don't understand the left side. You are taking the square-root of the numerator and denominator, but how does the square-root of the numerator simplify to 1?
    $endgroup$
    – Antoninus
    Dec 29 '18 at 16:48










  • $begingroup$
    $$frac{sqrt n}{n} = frac{1}{sqrt n}$$
    $endgroup$
    – KM101
    Dec 29 '18 at 16:51










  • $begingroup$
    Write $$frac{(9-x^2)^{1/2}}{(9-x^2)^{2/2}}$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 29 '18 at 16:52










  • $begingroup$
    Thank you, I've got it now!
    $endgroup$
    – Antoninus
    Dec 29 '18 at 17:11





















0












$begingroup$

Your first step is incorrect.



$$frac{a+b^{-n}}{c} color{red}{neq frac{a}{b^n+c}}$$



You can only bring the negative exponents down when the term is a factor of the whole numerator, not a factor of a single term in the numerator, as in



$$frac{ab^{-n}}{c} = frac{a}{b^nc}$$



Instead, you can make the substitution $u = 9-x^2$:



$$frac{u^{frac{1}{2}} + x^2(u)^{-frac{1}{2}}}{u} = frac{u^{frac{1}{2}}(1+x^2u^{-1})}{u} = frac{1+x^2u^{-1}}{u^{frac{1}{2}}} = frac{1}{u^{frac{1}{2}}}+frac{x^2u^{-1}}{u^{frac{1}{2}}} = frac{1}{u^{frac{1}{2}}}+frac{x^2}{ucdot u^{frac{1}{2}}}$$



Plugging in $u = 9-x^2$ and simplifying gives



$$frac{1}{(9-x^2)^{frac{1}{2}}}+frac{x^2}{(9-x^2)(9-x^2)^{frac{1}{2}}} = frac{9-x^2}{(9-x^2)(9-x^2)^{frac{1}{2}}}+frac{x^2}{(9-x^2)(9-x^2)^{frac{1}{2}}} = frac{9}{(9-x^2)^{frac{3}{2}}}$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Consider the substitution $z = sqrt{9 - x^2}$ and solve for $x$ and use the $z$ variable.






    share|cite|improve this answer









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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      Hint: Write $$frac{(9-x^2)^{1/2}}{9-x^2}+frac{x^2(9-x^2)^{-1/2}}{9-x^2}$$ and this is
      $$frac{1}{sqrt{9-x^2}}+frac{x^2}{(9-x^2)^{3/2}}$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Thank you for the feedback. I understand the right side of this, but I don't understand the left side. You are taking the square-root of the numerator and denominator, but how does the square-root of the numerator simplify to 1?
        $endgroup$
        – Antoninus
        Dec 29 '18 at 16:48










      • $begingroup$
        $$frac{sqrt n}{n} = frac{1}{sqrt n}$$
        $endgroup$
        – KM101
        Dec 29 '18 at 16:51










      • $begingroup$
        Write $$frac{(9-x^2)^{1/2}}{(9-x^2)^{2/2}}$$
        $endgroup$
        – Dr. Sonnhard Graubner
        Dec 29 '18 at 16:52










      • $begingroup$
        Thank you, I've got it now!
        $endgroup$
        – Antoninus
        Dec 29 '18 at 17:11


















      0












      $begingroup$

      Hint: Write $$frac{(9-x^2)^{1/2}}{9-x^2}+frac{x^2(9-x^2)^{-1/2}}{9-x^2}$$ and this is
      $$frac{1}{sqrt{9-x^2}}+frac{x^2}{(9-x^2)^{3/2}}$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Thank you for the feedback. I understand the right side of this, but I don't understand the left side. You are taking the square-root of the numerator and denominator, but how does the square-root of the numerator simplify to 1?
        $endgroup$
        – Antoninus
        Dec 29 '18 at 16:48










      • $begingroup$
        $$frac{sqrt n}{n} = frac{1}{sqrt n}$$
        $endgroup$
        – KM101
        Dec 29 '18 at 16:51










      • $begingroup$
        Write $$frac{(9-x^2)^{1/2}}{(9-x^2)^{2/2}}$$
        $endgroup$
        – Dr. Sonnhard Graubner
        Dec 29 '18 at 16:52










      • $begingroup$
        Thank you, I've got it now!
        $endgroup$
        – Antoninus
        Dec 29 '18 at 17:11
















      0












      0








      0





      $begingroup$

      Hint: Write $$frac{(9-x^2)^{1/2}}{9-x^2}+frac{x^2(9-x^2)^{-1/2}}{9-x^2}$$ and this is
      $$frac{1}{sqrt{9-x^2}}+frac{x^2}{(9-x^2)^{3/2}}$$






      share|cite|improve this answer









      $endgroup$



      Hint: Write $$frac{(9-x^2)^{1/2}}{9-x^2}+frac{x^2(9-x^2)^{-1/2}}{9-x^2}$$ and this is
      $$frac{1}{sqrt{9-x^2}}+frac{x^2}{(9-x^2)^{3/2}}$$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 29 '18 at 16:30









      Dr. Sonnhard GraubnerDr. Sonnhard Graubner

      79.5k42867




      79.5k42867












      • $begingroup$
        Thank you for the feedback. I understand the right side of this, but I don't understand the left side. You are taking the square-root of the numerator and denominator, but how does the square-root of the numerator simplify to 1?
        $endgroup$
        – Antoninus
        Dec 29 '18 at 16:48










      • $begingroup$
        $$frac{sqrt n}{n} = frac{1}{sqrt n}$$
        $endgroup$
        – KM101
        Dec 29 '18 at 16:51










      • $begingroup$
        Write $$frac{(9-x^2)^{1/2}}{(9-x^2)^{2/2}}$$
        $endgroup$
        – Dr. Sonnhard Graubner
        Dec 29 '18 at 16:52










      • $begingroup$
        Thank you, I've got it now!
        $endgroup$
        – Antoninus
        Dec 29 '18 at 17:11




















      • $begingroup$
        Thank you for the feedback. I understand the right side of this, but I don't understand the left side. You are taking the square-root of the numerator and denominator, but how does the square-root of the numerator simplify to 1?
        $endgroup$
        – Antoninus
        Dec 29 '18 at 16:48










      • $begingroup$
        $$frac{sqrt n}{n} = frac{1}{sqrt n}$$
        $endgroup$
        – KM101
        Dec 29 '18 at 16:51










      • $begingroup$
        Write $$frac{(9-x^2)^{1/2}}{(9-x^2)^{2/2}}$$
        $endgroup$
        – Dr. Sonnhard Graubner
        Dec 29 '18 at 16:52










      • $begingroup$
        Thank you, I've got it now!
        $endgroup$
        – Antoninus
        Dec 29 '18 at 17:11


















      $begingroup$
      Thank you for the feedback. I understand the right side of this, but I don't understand the left side. You are taking the square-root of the numerator and denominator, but how does the square-root of the numerator simplify to 1?
      $endgroup$
      – Antoninus
      Dec 29 '18 at 16:48




      $begingroup$
      Thank you for the feedback. I understand the right side of this, but I don't understand the left side. You are taking the square-root of the numerator and denominator, but how does the square-root of the numerator simplify to 1?
      $endgroup$
      – Antoninus
      Dec 29 '18 at 16:48












      $begingroup$
      $$frac{sqrt n}{n} = frac{1}{sqrt n}$$
      $endgroup$
      – KM101
      Dec 29 '18 at 16:51




      $begingroup$
      $$frac{sqrt n}{n} = frac{1}{sqrt n}$$
      $endgroup$
      – KM101
      Dec 29 '18 at 16:51












      $begingroup$
      Write $$frac{(9-x^2)^{1/2}}{(9-x^2)^{2/2}}$$
      $endgroup$
      – Dr. Sonnhard Graubner
      Dec 29 '18 at 16:52




      $begingroup$
      Write $$frac{(9-x^2)^{1/2}}{(9-x^2)^{2/2}}$$
      $endgroup$
      – Dr. Sonnhard Graubner
      Dec 29 '18 at 16:52












      $begingroup$
      Thank you, I've got it now!
      $endgroup$
      – Antoninus
      Dec 29 '18 at 17:11






      $begingroup$
      Thank you, I've got it now!
      $endgroup$
      – Antoninus
      Dec 29 '18 at 17:11













      0












      $begingroup$

      Your first step is incorrect.



      $$frac{a+b^{-n}}{c} color{red}{neq frac{a}{b^n+c}}$$



      You can only bring the negative exponents down when the term is a factor of the whole numerator, not a factor of a single term in the numerator, as in



      $$frac{ab^{-n}}{c} = frac{a}{b^nc}$$



      Instead, you can make the substitution $u = 9-x^2$:



      $$frac{u^{frac{1}{2}} + x^2(u)^{-frac{1}{2}}}{u} = frac{u^{frac{1}{2}}(1+x^2u^{-1})}{u} = frac{1+x^2u^{-1}}{u^{frac{1}{2}}} = frac{1}{u^{frac{1}{2}}}+frac{x^2u^{-1}}{u^{frac{1}{2}}} = frac{1}{u^{frac{1}{2}}}+frac{x^2}{ucdot u^{frac{1}{2}}}$$



      Plugging in $u = 9-x^2$ and simplifying gives



      $$frac{1}{(9-x^2)^{frac{1}{2}}}+frac{x^2}{(9-x^2)(9-x^2)^{frac{1}{2}}} = frac{9-x^2}{(9-x^2)(9-x^2)^{frac{1}{2}}}+frac{x^2}{(9-x^2)(9-x^2)^{frac{1}{2}}} = frac{9}{(9-x^2)^{frac{3}{2}}}$$






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        Your first step is incorrect.



        $$frac{a+b^{-n}}{c} color{red}{neq frac{a}{b^n+c}}$$



        You can only bring the negative exponents down when the term is a factor of the whole numerator, not a factor of a single term in the numerator, as in



        $$frac{ab^{-n}}{c} = frac{a}{b^nc}$$



        Instead, you can make the substitution $u = 9-x^2$:



        $$frac{u^{frac{1}{2}} + x^2(u)^{-frac{1}{2}}}{u} = frac{u^{frac{1}{2}}(1+x^2u^{-1})}{u} = frac{1+x^2u^{-1}}{u^{frac{1}{2}}} = frac{1}{u^{frac{1}{2}}}+frac{x^2u^{-1}}{u^{frac{1}{2}}} = frac{1}{u^{frac{1}{2}}}+frac{x^2}{ucdot u^{frac{1}{2}}}$$



        Plugging in $u = 9-x^2$ and simplifying gives



        $$frac{1}{(9-x^2)^{frac{1}{2}}}+frac{x^2}{(9-x^2)(9-x^2)^{frac{1}{2}}} = frac{9-x^2}{(9-x^2)(9-x^2)^{frac{1}{2}}}+frac{x^2}{(9-x^2)(9-x^2)^{frac{1}{2}}} = frac{9}{(9-x^2)^{frac{3}{2}}}$$






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Your first step is incorrect.



          $$frac{a+b^{-n}}{c} color{red}{neq frac{a}{b^n+c}}$$



          You can only bring the negative exponents down when the term is a factor of the whole numerator, not a factor of a single term in the numerator, as in



          $$frac{ab^{-n}}{c} = frac{a}{b^nc}$$



          Instead, you can make the substitution $u = 9-x^2$:



          $$frac{u^{frac{1}{2}} + x^2(u)^{-frac{1}{2}}}{u} = frac{u^{frac{1}{2}}(1+x^2u^{-1})}{u} = frac{1+x^2u^{-1}}{u^{frac{1}{2}}} = frac{1}{u^{frac{1}{2}}}+frac{x^2u^{-1}}{u^{frac{1}{2}}} = frac{1}{u^{frac{1}{2}}}+frac{x^2}{ucdot u^{frac{1}{2}}}$$



          Plugging in $u = 9-x^2$ and simplifying gives



          $$frac{1}{(9-x^2)^{frac{1}{2}}}+frac{x^2}{(9-x^2)(9-x^2)^{frac{1}{2}}} = frac{9-x^2}{(9-x^2)(9-x^2)^{frac{1}{2}}}+frac{x^2}{(9-x^2)(9-x^2)^{frac{1}{2}}} = frac{9}{(9-x^2)^{frac{3}{2}}}$$






          share|cite|improve this answer









          $endgroup$



          Your first step is incorrect.



          $$frac{a+b^{-n}}{c} color{red}{neq frac{a}{b^n+c}}$$



          You can only bring the negative exponents down when the term is a factor of the whole numerator, not a factor of a single term in the numerator, as in



          $$frac{ab^{-n}}{c} = frac{a}{b^nc}$$



          Instead, you can make the substitution $u = 9-x^2$:



          $$frac{u^{frac{1}{2}} + x^2(u)^{-frac{1}{2}}}{u} = frac{u^{frac{1}{2}}(1+x^2u^{-1})}{u} = frac{1+x^2u^{-1}}{u^{frac{1}{2}}} = frac{1}{u^{frac{1}{2}}}+frac{x^2u^{-1}}{u^{frac{1}{2}}} = frac{1}{u^{frac{1}{2}}}+frac{x^2}{ucdot u^{frac{1}{2}}}$$



          Plugging in $u = 9-x^2$ and simplifying gives



          $$frac{1}{(9-x^2)^{frac{1}{2}}}+frac{x^2}{(9-x^2)(9-x^2)^{frac{1}{2}}} = frac{9-x^2}{(9-x^2)(9-x^2)^{frac{1}{2}}}+frac{x^2}{(9-x^2)(9-x^2)^{frac{1}{2}}} = frac{9}{(9-x^2)^{frac{3}{2}}}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 29 '18 at 16:49









          KM101KM101

          6,0861525




          6,0861525























              0












              $begingroup$

              Consider the substitution $z = sqrt{9 - x^2}$ and solve for $x$ and use the $z$ variable.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Consider the substitution $z = sqrt{9 - x^2}$ and solve for $x$ and use the $z$ variable.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Consider the substitution $z = sqrt{9 - x^2}$ and solve for $x$ and use the $z$ variable.






                  share|cite|improve this answer









                  $endgroup$



                  Consider the substitution $z = sqrt{9 - x^2}$ and solve for $x$ and use the $z$ variable.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 29 '18 at 19:12









                  Lorenz H MenkeLorenz H Menke

                  9611




                  9611






























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                      Verónica Boquete

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                      Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
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                      Ida-Boy-Ed-Garten

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                      Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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