Quotient rule of radicals - cannot get to correct answer












1












$begingroup$


I am to simplify the following expression:



$$dfrac{sqrt{(9a^5b^{14})}}{sqrt{(3a^4b^5)}}$$



The solution is given:



$$b^4sqrt{3ab}$$



I was unable to recreate this solution. Here's where I got to. Using the quotient rule I rewrote the expression as:



$$sqrt{dfrac{9a^5b^{14}}{3a^4b^5}}$$



Then, I attempted to simplify (the radicand?):



$$dfrac{9a^5b^{14}}{3a^4b^5}$$



Becomes $sqrt{3ab^9}$.



My train of thought is that first, I simplify $9a^5 / 3a^4$ to just $3a$. Is that right?



Next I tried to simplify $b^{14} / b^5$ to $b^9$



Thus I arrived at $sqrt{3ab^9}$



Where did I go wrong and am I on the right track?






algebra-precalculus







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    You can use curly brackets to include multiple characters within a radical.
    $endgroup$
    – KM101
    Dec 31 '18 at 17:42












  • $begingroup$
    Thanks for the tip, added now
    $endgroup$
    – Doug Fir
    Dec 31 '18 at 17:48






  • 1




    $begingroup$
    No problem. Also, you can also use frac{}{} to show fractions, such as $frac{a}{b}$, which is typically preferred to using a slash, like $a/b$. (Of course, you don’t need to edit the post. Just pointing out.)
    $endgroup$
    – KM101
    Dec 31 '18 at 17:50










  • $begingroup$
    Noted with thanks!
    $endgroup$
    – Doug Fir
    Dec 31 '18 at 17:50
















1












$begingroup$


I am to simplify the following expression:



$$dfrac{sqrt{(9a^5b^{14})}}{sqrt{(3a^4b^5)}}$$



The solution is given:



$$b^4sqrt{3ab}$$



I was unable to recreate this solution. Here's where I got to. Using the quotient rule I rewrote the expression as:



$$sqrt{dfrac{9a^5b^{14}}{3a^4b^5}}$$



Then, I attempted to simplify (the radicand?):



$$dfrac{9a^5b^{14}}{3a^4b^5}$$



Becomes $sqrt{3ab^9}$.



My train of thought is that first, I simplify $9a^5 / 3a^4$ to just $3a$. Is that right?



Next I tried to simplify $b^{14} / b^5$ to $b^9$



Thus I arrived at $sqrt{3ab^9}$



Where did I go wrong and am I on the right track?






algebra-precalculus







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    You can use curly brackets to include multiple characters within a radical.
    $endgroup$
    – KM101
    Dec 31 '18 at 17:42












  • $begingroup$
    Thanks for the tip, added now
    $endgroup$
    – Doug Fir
    Dec 31 '18 at 17:48






  • 1




    $begingroup$
    No problem. Also, you can also use frac{}{} to show fractions, such as $frac{a}{b}$, which is typically preferred to using a slash, like $a/b$. (Of course, you don’t need to edit the post. Just pointing out.)
    $endgroup$
    – KM101
    Dec 31 '18 at 17:50










  • $begingroup$
    Noted with thanks!
    $endgroup$
    – Doug Fir
    Dec 31 '18 at 17:50














1












1








1





$begingroup$


I am to simplify the following expression:



$$dfrac{sqrt{(9a^5b^{14})}}{sqrt{(3a^4b^5)}}$$



The solution is given:



$$b^4sqrt{3ab}$$



I was unable to recreate this solution. Here's where I got to. Using the quotient rule I rewrote the expression as:



$$sqrt{dfrac{9a^5b^{14}}{3a^4b^5}}$$



Then, I attempted to simplify (the radicand?):



$$dfrac{9a^5b^{14}}{3a^4b^5}$$



Becomes $sqrt{3ab^9}$.



My train of thought is that first, I simplify $9a^5 / 3a^4$ to just $3a$. Is that right?



Next I tried to simplify $b^{14} / b^5$ to $b^9$



Thus I arrived at $sqrt{3ab^9}$



Where did I go wrong and am I on the right track?






algebra-precalculus







share|cite|improve this question











$endgroup$




I am to simplify the following expression:



$$dfrac{sqrt{(9a^5b^{14})}}{sqrt{(3a^4b^5)}}$$



The solution is given:



$$b^4sqrt{3ab}$$



I was unable to recreate this solution. Here's where I got to. Using the quotient rule I rewrote the expression as:



$$sqrt{dfrac{9a^5b^{14}}{3a^4b^5}}$$



Then, I attempted to simplify (the radicand?):



$$dfrac{9a^5b^{14}}{3a^4b^5}$$



Becomes $sqrt{3ab^9}$.



My train of thought is that first, I simplify $9a^5 / 3a^4$ to just $3a$. Is that right?



Next I tried to simplify $b^{14} / b^5$ to $b^9$



Thus I arrived at $sqrt{3ab^9}$



Where did I go wrong and am I on the right track?






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 17:56









steven gregory

18.2k32258




18.2k32258










asked Dec 31 '18 at 17:38









Doug FirDoug Fir

3828




3828








  • 2




    $begingroup$
    You can use curly brackets to include multiple characters within a radical.
    $endgroup$
    – KM101
    Dec 31 '18 at 17:42












  • $begingroup$
    Thanks for the tip, added now
    $endgroup$
    – Doug Fir
    Dec 31 '18 at 17:48






  • 1




    $begingroup$
    No problem. Also, you can also use frac{}{} to show fractions, such as $frac{a}{b}$, which is typically preferred to using a slash, like $a/b$. (Of course, you don’t need to edit the post. Just pointing out.)
    $endgroup$
    – KM101
    Dec 31 '18 at 17:50










  • $begingroup$
    Noted with thanks!
    $endgroup$
    – Doug Fir
    Dec 31 '18 at 17:50














  • 2




    $begingroup$
    You can use curly brackets to include multiple characters within a radical.
    $endgroup$
    – KM101
    Dec 31 '18 at 17:42












  • $begingroup$
    Thanks for the tip, added now
    $endgroup$
    – Doug Fir
    Dec 31 '18 at 17:48






  • 1




    $begingroup$
    No problem. Also, you can also use frac{}{} to show fractions, such as $frac{a}{b}$, which is typically preferred to using a slash, like $a/b$. (Of course, you don’t need to edit the post. Just pointing out.)
    $endgroup$
    – KM101
    Dec 31 '18 at 17:50










  • $begingroup$
    Noted with thanks!
    $endgroup$
    – Doug Fir
    Dec 31 '18 at 17:50








2




2




$begingroup$
You can use curly brackets to include multiple characters within a radical.
$endgroup$
– KM101
Dec 31 '18 at 17:42






$begingroup$
You can use curly brackets to include multiple characters within a radical.
$endgroup$
– KM101
Dec 31 '18 at 17:42














$begingroup$
Thanks for the tip, added now
$endgroup$
– Doug Fir
Dec 31 '18 at 17:48




$begingroup$
Thanks for the tip, added now
$endgroup$
– Doug Fir
Dec 31 '18 at 17:48




1




1




$begingroup$
No problem. Also, you can also use frac{}{} to show fractions, such as $frac{a}{b}$, which is typically preferred to using a slash, like $a/b$. (Of course, you don’t need to edit the post. Just pointing out.)
$endgroup$
– KM101
Dec 31 '18 at 17:50




$begingroup$
No problem. Also, you can also use frac{}{} to show fractions, such as $frac{a}{b}$, which is typically preferred to using a slash, like $a/b$. (Of course, you don’t need to edit the post. Just pointing out.)
$endgroup$
– KM101
Dec 31 '18 at 17:50












$begingroup$
Noted with thanks!
$endgroup$
– Doug Fir
Dec 31 '18 at 17:50




$begingroup$
Noted with thanks!
$endgroup$
– Doug Fir
Dec 31 '18 at 17:50










3 Answers
3






active

oldest

votes


















2












$begingroup$

Notice that



$$sqrt{3ab^9}=(sqrt{3a})(b^frac92)=b^4sqrt{3ab}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for answering but I'm struggling to follow. Why $b^{9/2}$?
    $endgroup$
    – Doug Fir
    Dec 31 '18 at 17:49






  • 1




    $begingroup$
    Because $sqrt b = b^{frac{1}{2}}$, so $sqrt{b^9} = b^{frac{9}{2}}$. In general, we have $sqrt[n] {b^m} = b^{frac{m}{n}}$.
    $endgroup$
    – KM101
    Dec 31 '18 at 17:50












  • $begingroup$
    Thank you but I still struggle to see. How did you complete the last step, to go from $b^{9/2}$ to $b^4$
    $endgroup$
    – Doug Fir
    Dec 31 '18 at 17:54






  • 1




    $begingroup$
    It’s the reverse process: $$b^{frac{9}{2}} = b^4cdot b^{frac{1}{2}} = b^4cdotsqrt{b}$$ In fact, as seen, your answer isn’t even wrong, it’s just not fully simplified.
    $endgroup$
    – KM101
    Dec 31 '18 at 17:55








  • 1




    $begingroup$
    @DougFir $b^9=b^8cdot b$, and $sqrt{b^8}=b^4$. So you can take $b^4$ outside the radical and leave the remaining $b$ inside.
    $endgroup$
    – timtfj
    Dec 31 '18 at 18:26





















3












$begingroup$

$b^9= (b^8)b= (b^4)^2b$ so $sqrt{b^9}= sqrt{(b^4)^2}sqrt{b}= b^4sqrt{b}$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Hint: Write $$sqrt{frac{9a^5b^{14}}{3a^4b^5}}=sqrt{3ab^9}=sqrt{3ab^9}=b^4sqrt{3ab}$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      $sqrt {3ab^9}=sqrt {3ab^9}?$
      $endgroup$
      – Mohammad Zuhair Khan
      Dec 31 '18 at 17:45






    • 2




      $begingroup$
      @MohammadZuhairKhan He is not wrong
      $endgroup$
      – cansomeonehelpmeout
      Dec 31 '18 at 17:52










    • $begingroup$
      @cansomeonehelpmeout I am forced to agree.
      $endgroup$
      – Mohammad Zuhair Khan
      Dec 31 '18 at 17:54











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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Notice that



    $$sqrt{3ab^9}=(sqrt{3a})(b^frac92)=b^4sqrt{3ab}$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you for answering but I'm struggling to follow. Why $b^{9/2}$?
      $endgroup$
      – Doug Fir
      Dec 31 '18 at 17:49






    • 1




      $begingroup$
      Because $sqrt b = b^{frac{1}{2}}$, so $sqrt{b^9} = b^{frac{9}{2}}$. In general, we have $sqrt[n] {b^m} = b^{frac{m}{n}}$.
      $endgroup$
      – KM101
      Dec 31 '18 at 17:50












    • $begingroup$
      Thank you but I still struggle to see. How did you complete the last step, to go from $b^{9/2}$ to $b^4$
      $endgroup$
      – Doug Fir
      Dec 31 '18 at 17:54






    • 1




      $begingroup$
      It’s the reverse process: $$b^{frac{9}{2}} = b^4cdot b^{frac{1}{2}} = b^4cdotsqrt{b}$$ In fact, as seen, your answer isn’t even wrong, it’s just not fully simplified.
      $endgroup$
      – KM101
      Dec 31 '18 at 17:55








    • 1




      $begingroup$
      @DougFir $b^9=b^8cdot b$, and $sqrt{b^8}=b^4$. So you can take $b^4$ outside the radical and leave the remaining $b$ inside.
      $endgroup$
      – timtfj
      Dec 31 '18 at 18:26


















    2












    $begingroup$

    Notice that



    $$sqrt{3ab^9}=(sqrt{3a})(b^frac92)=b^4sqrt{3ab}$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you for answering but I'm struggling to follow. Why $b^{9/2}$?
      $endgroup$
      – Doug Fir
      Dec 31 '18 at 17:49






    • 1




      $begingroup$
      Because $sqrt b = b^{frac{1}{2}}$, so $sqrt{b^9} = b^{frac{9}{2}}$. In general, we have $sqrt[n] {b^m} = b^{frac{m}{n}}$.
      $endgroup$
      – KM101
      Dec 31 '18 at 17:50












    • $begingroup$
      Thank you but I still struggle to see. How did you complete the last step, to go from $b^{9/2}$ to $b^4$
      $endgroup$
      – Doug Fir
      Dec 31 '18 at 17:54






    • 1




      $begingroup$
      It’s the reverse process: $$b^{frac{9}{2}} = b^4cdot b^{frac{1}{2}} = b^4cdotsqrt{b}$$ In fact, as seen, your answer isn’t even wrong, it’s just not fully simplified.
      $endgroup$
      – KM101
      Dec 31 '18 at 17:55








    • 1




      $begingroup$
      @DougFir $b^9=b^8cdot b$, and $sqrt{b^8}=b^4$. So you can take $b^4$ outside the radical and leave the remaining $b$ inside.
      $endgroup$
      – timtfj
      Dec 31 '18 at 18:26
















    2












    2








    2





    $begingroup$

    Notice that



    $$sqrt{3ab^9}=(sqrt{3a})(b^frac92)=b^4sqrt{3ab}$$






    share|cite|improve this answer









    $endgroup$



    Notice that



    $$sqrt{3ab^9}=(sqrt{3a})(b^frac92)=b^4sqrt{3ab}$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 31 '18 at 17:42









    Siong Thye GohSiong Thye Goh

    102k1466118




    102k1466118












    • $begingroup$
      Thank you for answering but I'm struggling to follow. Why $b^{9/2}$?
      $endgroup$
      – Doug Fir
      Dec 31 '18 at 17:49






    • 1




      $begingroup$
      Because $sqrt b = b^{frac{1}{2}}$, so $sqrt{b^9} = b^{frac{9}{2}}$. In general, we have $sqrt[n] {b^m} = b^{frac{m}{n}}$.
      $endgroup$
      – KM101
      Dec 31 '18 at 17:50












    • $begingroup$
      Thank you but I still struggle to see. How did you complete the last step, to go from $b^{9/2}$ to $b^4$
      $endgroup$
      – Doug Fir
      Dec 31 '18 at 17:54






    • 1




      $begingroup$
      It’s the reverse process: $$b^{frac{9}{2}} = b^4cdot b^{frac{1}{2}} = b^4cdotsqrt{b}$$ In fact, as seen, your answer isn’t even wrong, it’s just not fully simplified.
      $endgroup$
      – KM101
      Dec 31 '18 at 17:55








    • 1




      $begingroup$
      @DougFir $b^9=b^8cdot b$, and $sqrt{b^8}=b^4$. So you can take $b^4$ outside the radical and leave the remaining $b$ inside.
      $endgroup$
      – timtfj
      Dec 31 '18 at 18:26




















    • $begingroup$
      Thank you for answering but I'm struggling to follow. Why $b^{9/2}$?
      $endgroup$
      – Doug Fir
      Dec 31 '18 at 17:49






    • 1




      $begingroup$
      Because $sqrt b = b^{frac{1}{2}}$, so $sqrt{b^9} = b^{frac{9}{2}}$. In general, we have $sqrt[n] {b^m} = b^{frac{m}{n}}$.
      $endgroup$
      – KM101
      Dec 31 '18 at 17:50












    • $begingroup$
      Thank you but I still struggle to see. How did you complete the last step, to go from $b^{9/2}$ to $b^4$
      $endgroup$
      – Doug Fir
      Dec 31 '18 at 17:54






    • 1




      $begingroup$
      It’s the reverse process: $$b^{frac{9}{2}} = b^4cdot b^{frac{1}{2}} = b^4cdotsqrt{b}$$ In fact, as seen, your answer isn’t even wrong, it’s just not fully simplified.
      $endgroup$
      – KM101
      Dec 31 '18 at 17:55








    • 1




      $begingroup$
      @DougFir $b^9=b^8cdot b$, and $sqrt{b^8}=b^4$. So you can take $b^4$ outside the radical and leave the remaining $b$ inside.
      $endgroup$
      – timtfj
      Dec 31 '18 at 18:26


















    $begingroup$
    Thank you for answering but I'm struggling to follow. Why $b^{9/2}$?
    $endgroup$
    – Doug Fir
    Dec 31 '18 at 17:49




    $begingroup$
    Thank you for answering but I'm struggling to follow. Why $b^{9/2}$?
    $endgroup$
    – Doug Fir
    Dec 31 '18 at 17:49




    1




    1




    $begingroup$
    Because $sqrt b = b^{frac{1}{2}}$, so $sqrt{b^9} = b^{frac{9}{2}}$. In general, we have $sqrt[n] {b^m} = b^{frac{m}{n}}$.
    $endgroup$
    – KM101
    Dec 31 '18 at 17:50






    $begingroup$
    Because $sqrt b = b^{frac{1}{2}}$, so $sqrt{b^9} = b^{frac{9}{2}}$. In general, we have $sqrt[n] {b^m} = b^{frac{m}{n}}$.
    $endgroup$
    – KM101
    Dec 31 '18 at 17:50














    $begingroup$
    Thank you but I still struggle to see. How did you complete the last step, to go from $b^{9/2}$ to $b^4$
    $endgroup$
    – Doug Fir
    Dec 31 '18 at 17:54




    $begingroup$
    Thank you but I still struggle to see. How did you complete the last step, to go from $b^{9/2}$ to $b^4$
    $endgroup$
    – Doug Fir
    Dec 31 '18 at 17:54




    1




    1




    $begingroup$
    It’s the reverse process: $$b^{frac{9}{2}} = b^4cdot b^{frac{1}{2}} = b^4cdotsqrt{b}$$ In fact, as seen, your answer isn’t even wrong, it’s just not fully simplified.
    $endgroup$
    – KM101
    Dec 31 '18 at 17:55






    $begingroup$
    It’s the reverse process: $$b^{frac{9}{2}} = b^4cdot b^{frac{1}{2}} = b^4cdotsqrt{b}$$ In fact, as seen, your answer isn’t even wrong, it’s just not fully simplified.
    $endgroup$
    – KM101
    Dec 31 '18 at 17:55






    1




    1




    $begingroup$
    @DougFir $b^9=b^8cdot b$, and $sqrt{b^8}=b^4$. So you can take $b^4$ outside the radical and leave the remaining $b$ inside.
    $endgroup$
    – timtfj
    Dec 31 '18 at 18:26






    $begingroup$
    @DougFir $b^9=b^8cdot b$, and $sqrt{b^8}=b^4$. So you can take $b^4$ outside the radical and leave the remaining $b$ inside.
    $endgroup$
    – timtfj
    Dec 31 '18 at 18:26













    3












    $begingroup$

    $b^9= (b^8)b= (b^4)^2b$ so $sqrt{b^9}= sqrt{(b^4)^2}sqrt{b}= b^4sqrt{b}$.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      $b^9= (b^8)b= (b^4)^2b$ so $sqrt{b^9}= sqrt{(b^4)^2}sqrt{b}= b^4sqrt{b}$.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        $b^9= (b^8)b= (b^4)^2b$ so $sqrt{b^9}= sqrt{(b^4)^2}sqrt{b}= b^4sqrt{b}$.






        share|cite|improve this answer









        $endgroup$



        $b^9= (b^8)b= (b^4)^2b$ so $sqrt{b^9}= sqrt{(b^4)^2}sqrt{b}= b^4sqrt{b}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 31 '18 at 17:51









        user247327user247327

        11.2k1515




        11.2k1515























            0












            $begingroup$

            Hint: Write $$sqrt{frac{9a^5b^{14}}{3a^4b^5}}=sqrt{3ab^9}=sqrt{3ab^9}=b^4sqrt{3ab}$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              $sqrt {3ab^9}=sqrt {3ab^9}?$
              $endgroup$
              – Mohammad Zuhair Khan
              Dec 31 '18 at 17:45






            • 2




              $begingroup$
              @MohammadZuhairKhan He is not wrong
              $endgroup$
              – cansomeonehelpmeout
              Dec 31 '18 at 17:52










            • $begingroup$
              @cansomeonehelpmeout I am forced to agree.
              $endgroup$
              – Mohammad Zuhair Khan
              Dec 31 '18 at 17:54
















            0












            $begingroup$

            Hint: Write $$sqrt{frac{9a^5b^{14}}{3a^4b^5}}=sqrt{3ab^9}=sqrt{3ab^9}=b^4sqrt{3ab}$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              $sqrt {3ab^9}=sqrt {3ab^9}?$
              $endgroup$
              – Mohammad Zuhair Khan
              Dec 31 '18 at 17:45






            • 2




              $begingroup$
              @MohammadZuhairKhan He is not wrong
              $endgroup$
              – cansomeonehelpmeout
              Dec 31 '18 at 17:52










            • $begingroup$
              @cansomeonehelpmeout I am forced to agree.
              $endgroup$
              – Mohammad Zuhair Khan
              Dec 31 '18 at 17:54














            0












            0








            0





            $begingroup$

            Hint: Write $$sqrt{frac{9a^5b^{14}}{3a^4b^5}}=sqrt{3ab^9}=sqrt{3ab^9}=b^4sqrt{3ab}$$






            share|cite|improve this answer











            $endgroup$



            Hint: Write $$sqrt{frac{9a^5b^{14}}{3a^4b^5}}=sqrt{3ab^9}=sqrt{3ab^9}=b^4sqrt{3ab}$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 31 '18 at 17:46

























            answered Dec 31 '18 at 17:42









            Dr. Sonnhard GraubnerDr. Sonnhard Graubner

            76.5k42866




            76.5k42866












            • $begingroup$
              $sqrt {3ab^9}=sqrt {3ab^9}?$
              $endgroup$
              – Mohammad Zuhair Khan
              Dec 31 '18 at 17:45






            • 2




              $begingroup$
              @MohammadZuhairKhan He is not wrong
              $endgroup$
              – cansomeonehelpmeout
              Dec 31 '18 at 17:52










            • $begingroup$
              @cansomeonehelpmeout I am forced to agree.
              $endgroup$
              – Mohammad Zuhair Khan
              Dec 31 '18 at 17:54


















            • $begingroup$
              $sqrt {3ab^9}=sqrt {3ab^9}?$
              $endgroup$
              – Mohammad Zuhair Khan
              Dec 31 '18 at 17:45






            • 2




              $begingroup$
              @MohammadZuhairKhan He is not wrong
              $endgroup$
              – cansomeonehelpmeout
              Dec 31 '18 at 17:52










            • $begingroup$
              @cansomeonehelpmeout I am forced to agree.
              $endgroup$
              – Mohammad Zuhair Khan
              Dec 31 '18 at 17:54
















            $begingroup$
            $sqrt {3ab^9}=sqrt {3ab^9}?$
            $endgroup$
            – Mohammad Zuhair Khan
            Dec 31 '18 at 17:45




            $begingroup$
            $sqrt {3ab^9}=sqrt {3ab^9}?$
            $endgroup$
            – Mohammad Zuhair Khan
            Dec 31 '18 at 17:45




            2




            2




            $begingroup$
            @MohammadZuhairKhan He is not wrong
            $endgroup$
            – cansomeonehelpmeout
            Dec 31 '18 at 17:52




            $begingroup$
            @MohammadZuhairKhan He is not wrong
            $endgroup$
            – cansomeonehelpmeout
            Dec 31 '18 at 17:52












            $begingroup$
            @cansomeonehelpmeout I am forced to agree.
            $endgroup$
            – Mohammad Zuhair Khan
            Dec 31 '18 at 17:54




            $begingroup$
            @cansomeonehelpmeout I am forced to agree.
            $endgroup$
            – Mohammad Zuhair Khan
            Dec 31 '18 at 17:54


















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