If $c,d$ are the real roots of the equation, $(x-a)(x-b)=f$

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EDIT:




Sorry, corrected the Typo




If $c,d$ are the real roots of the equation,
$$(x-a)(x-b)=f$$
Then roots of the equation, $$(x-c)(x-d)+f=0$$
are?



  1. $a,b$

  2. $fracaf,fracbf$

  3. $fracfa,fracfb$

  4. $fraccf,fracdf$

I don't have any hints on how to proceed with this. The only thing I was able to establish was, $$a+b=c+d$$



$$ab=cd+f$$



Any hints would be helpful. Thank you.




systems-of-equations quadratics




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  • 2




    Did you mean $(x-c)(x-d)+f=0$?
    – lulu
    Aug 2 at 12:42






  • 1




    $(x-c)+(x-d)+f=0$ has only a solution. Are you sure this is the correct equation?
    – mfl
    Aug 2 at 12:42










  • You have $(x-a)(x-b)-f = (x-c)(x-d)$
    – Hari Shankar
    Aug 2 at 12:47














up vote
1
down vote

favorite












EDIT:




Sorry, corrected the Typo




If $c,d$ are the real roots of the equation,
$$(x-a)(x-b)=f$$
Then roots of the equation, $$(x-c)(x-d)+f=0$$
are?



  1. $a,b$

  2. $fracaf,fracbf$

  3. $fracfa,fracfb$

  4. $fraccf,fracdf$

I don't have any hints on how to proceed with this. The only thing I was able to establish was, $$a+b=c+d$$



$$ab=cd+f$$



Any hints would be helpful. Thank you.




systems-of-equations quadratics




share|cite|improve this question

















  • 2




    Did you mean $(x-c)(x-d)+f=0$?
    – lulu
    Aug 2 at 12:42






  • 1




    $(x-c)+(x-d)+f=0$ has only a solution. Are you sure this is the correct equation?
    – mfl
    Aug 2 at 12:42










  • You have $(x-a)(x-b)-f = (x-c)(x-d)$
    – Hari Shankar
    Aug 2 at 12:47












up vote
1
down vote

favorite









up vote
1
down vote

favorite











EDIT:




Sorry, corrected the Typo




If $c,d$ are the real roots of the equation,
$$(x-a)(x-b)=f$$
Then roots of the equation, $$(x-c)(x-d)+f=0$$
are?



  1. $a,b$

  2. $fracaf,fracbf$

  3. $fracfa,fracfb$

  4. $fraccf,fracdf$

I don't have any hints on how to proceed with this. The only thing I was able to establish was, $$a+b=c+d$$



$$ab=cd+f$$



Any hints would be helpful. Thank you.




systems-of-equations quadratics




share|cite|improve this question













EDIT:




Sorry, corrected the Typo




If $c,d$ are the real roots of the equation,
$$(x-a)(x-b)=f$$
Then roots of the equation, $$(x-c)(x-d)+f=0$$
are?



  1. $a,b$

  2. $fracaf,fracbf$

  3. $fracfa,fracfb$

  4. $fraccf,fracdf$

I don't have any hints on how to proceed with this. The only thing I was able to establish was, $$a+b=c+d$$



$$ab=cd+f$$



Any hints would be helpful. Thank you.





systems-of-equations quadratics





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 2 at 13:06
























asked Aug 2 at 12:40









prog_SAHIL

773217




773217







  • 2




    Did you mean $(x-c)(x-d)+f=0$?
    – lulu
    Aug 2 at 12:42






  • 1




    $(x-c)+(x-d)+f=0$ has only a solution. Are you sure this is the correct equation?
    – mfl
    Aug 2 at 12:42










  • You have $(x-a)(x-b)-f = (x-c)(x-d)$
    – Hari Shankar
    Aug 2 at 12:47












  • 2




    Did you mean $(x-c)(x-d)+f=0$?
    – lulu
    Aug 2 at 12:42






  • 1




    $(x-c)+(x-d)+f=0$ has only a solution. Are you sure this is the correct equation?
    – mfl
    Aug 2 at 12:42










  • You have $(x-a)(x-b)-f = (x-c)(x-d)$
    – Hari Shankar
    Aug 2 at 12:47







2




2




Did you mean $(x-c)(x-d)+f=0$?
– lulu
Aug 2 at 12:42




Did you mean $(x-c)(x-d)+f=0$?
– lulu
Aug 2 at 12:42




1




1




$(x-c)+(x-d)+f=0$ has only a solution. Are you sure this is the correct equation?
– mfl
Aug 2 at 12:42




$(x-c)+(x-d)+f=0$ has only a solution. Are you sure this is the correct equation?
– mfl
Aug 2 at 12:42












You have $(x-a)(x-b)-f = (x-c)(x-d)$
– Hari Shankar
Aug 2 at 12:47




You have $(x-a)(x-b)-f = (x-c)(x-d)$
– Hari Shankar
Aug 2 at 12:47










2 Answers
2






active

oldest

votes

















up vote
2
down vote



accepted










The answer can be achieved by using the same technique again.



That is, consider the roots of the equation
$$(x-a)(x-b)-f = x^2 - (a+b)x +(ab-f) = 0$$



As the OP has already realised, the sum of the roots of this quadratic are $(a+b)$, and the product is $(ab-f)$.



Given that the roots are $c,d$, we can thus conclude that
$$ c+d = a+b quad cd = ab-f $$



We now simply apply this same logic to the second equation.
$$ (x-c)(x-d) + f = 0 implies x^2-(c+d)x +(cd-f) = 0$$



So the sum of the roots of this equation are $c+d$ and the product of the roots are ($cd-f)$.



But we have already determined that $$ c+d = a+b quad cd = ab-f $$



Thus, the sum roots of the second equation are $a+b$, and the product is $ab$.



Thus it is clear that the answer is option 1: the roots are $a,b$






share|cite|improve this answer























  • It was a typo in the textbook. That's what had me stuck. Thank you.
    – prog_SAHIL
    Aug 2 at 13:07

















up vote
2
down vote













Trusting that $(x-c)(x-d)+f=0$ was intended, just remark that $$(x-a)(x-b)-f=(x-c)(x-d)implies (x-c)(x-d)+f=(x-a)(x-b)$$



Easy to see that this generalizes to higher degree.






share|cite|improve this answer





















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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    2
    down vote



    accepted










    The answer can be achieved by using the same technique again.



    That is, consider the roots of the equation
    $$(x-a)(x-b)-f = x^2 - (a+b)x +(ab-f) = 0$$



    As the OP has already realised, the sum of the roots of this quadratic are $(a+b)$, and the product is $(ab-f)$.



    Given that the roots are $c,d$, we can thus conclude that
    $$ c+d = a+b quad cd = ab-f $$



    We now simply apply this same logic to the second equation.
    $$ (x-c)(x-d) + f = 0 implies x^2-(c+d)x +(cd-f) = 0$$



    So the sum of the roots of this equation are $c+d$ and the product of the roots are ($cd-f)$.



    But we have already determined that $$ c+d = a+b quad cd = ab-f $$



    Thus, the sum roots of the second equation are $a+b$, and the product is $ab$.



    Thus it is clear that the answer is option 1: the roots are $a,b$






    share|cite|improve this answer























    • It was a typo in the textbook. That's what had me stuck. Thank you.
      – prog_SAHIL
      Aug 2 at 13:07














    up vote
    2
    down vote



    accepted










    The answer can be achieved by using the same technique again.



    That is, consider the roots of the equation
    $$(x-a)(x-b)-f = x^2 - (a+b)x +(ab-f) = 0$$



    As the OP has already realised, the sum of the roots of this quadratic are $(a+b)$, and the product is $(ab-f)$.



    Given that the roots are $c,d$, we can thus conclude that
    $$ c+d = a+b quad cd = ab-f $$



    We now simply apply this same logic to the second equation.
    $$ (x-c)(x-d) + f = 0 implies x^2-(c+d)x +(cd-f) = 0$$



    So the sum of the roots of this equation are $c+d$ and the product of the roots are ($cd-f)$.



    But we have already determined that $$ c+d = a+b quad cd = ab-f $$



    Thus, the sum roots of the second equation are $a+b$, and the product is $ab$.



    Thus it is clear that the answer is option 1: the roots are $a,b$






    share|cite|improve this answer























    • It was a typo in the textbook. That's what had me stuck. Thank you.
      – prog_SAHIL
      Aug 2 at 13:07












    up vote
    2
    down vote



    accepted







    up vote
    2
    down vote



    accepted






    The answer can be achieved by using the same technique again.



    That is, consider the roots of the equation
    $$(x-a)(x-b)-f = x^2 - (a+b)x +(ab-f) = 0$$



    As the OP has already realised, the sum of the roots of this quadratic are $(a+b)$, and the product is $(ab-f)$.



    Given that the roots are $c,d$, we can thus conclude that
    $$ c+d = a+b quad cd = ab-f $$



    We now simply apply this same logic to the second equation.
    $$ (x-c)(x-d) + f = 0 implies x^2-(c+d)x +(cd-f) = 0$$



    So the sum of the roots of this equation are $c+d$ and the product of the roots are ($cd-f)$.



    But we have already determined that $$ c+d = a+b quad cd = ab-f $$



    Thus, the sum roots of the second equation are $a+b$, and the product is $ab$.



    Thus it is clear that the answer is option 1: the roots are $a,b$






    share|cite|improve this answer















    The answer can be achieved by using the same technique again.



    That is, consider the roots of the equation
    $$(x-a)(x-b)-f = x^2 - (a+b)x +(ab-f) = 0$$



    As the OP has already realised, the sum of the roots of this quadratic are $(a+b)$, and the product is $(ab-f)$.



    Given that the roots are $c,d$, we can thus conclude that
    $$ c+d = a+b quad cd = ab-f $$



    We now simply apply this same logic to the second equation.
    $$ (x-c)(x-d) + f = 0 implies x^2-(c+d)x +(cd-f) = 0$$



    So the sum of the roots of this equation are $c+d$ and the product of the roots are ($cd-f)$.



    But we have already determined that $$ c+d = a+b quad cd = ab-f $$



    Thus, the sum roots of the second equation are $a+b$, and the product is $ab$.



    Thus it is clear that the answer is option 1: the roots are $a,b$







    share|cite|improve this answer















    share|cite|improve this answer



    share|cite|improve this answer








    edited Aug 2 at 21:35


























    answered Aug 2 at 12:54









    Martin Roberts

    1,204318




    1,204318











    • It was a typo in the textbook. That's what had me stuck. Thank you.
      – prog_SAHIL
      Aug 2 at 13:07
















    • It was a typo in the textbook. That's what had me stuck. Thank you.
      – prog_SAHIL
      Aug 2 at 13:07















    It was a typo in the textbook. That's what had me stuck. Thank you.
    – prog_SAHIL
    Aug 2 at 13:07




    It was a typo in the textbook. That's what had me stuck. Thank you.
    – prog_SAHIL
    Aug 2 at 13:07










    up vote
    2
    down vote













    Trusting that $(x-c)(x-d)+f=0$ was intended, just remark that $$(x-a)(x-b)-f=(x-c)(x-d)implies (x-c)(x-d)+f=(x-a)(x-b)$$



    Easy to see that this generalizes to higher degree.






    share|cite|improve this answer

























      up vote
      2
      down vote













      Trusting that $(x-c)(x-d)+f=0$ was intended, just remark that $$(x-a)(x-b)-f=(x-c)(x-d)implies (x-c)(x-d)+f=(x-a)(x-b)$$



      Easy to see that this generalizes to higher degree.






      share|cite|improve this answer























        up vote
        2
        down vote










        up vote
        2
        down vote









        Trusting that $(x-c)(x-d)+f=0$ was intended, just remark that $$(x-a)(x-b)-f=(x-c)(x-d)implies (x-c)(x-d)+f=(x-a)(x-b)$$



        Easy to see that this generalizes to higher degree.






        share|cite|improve this answer













        Trusting that $(x-c)(x-d)+f=0$ was intended, just remark that $$(x-a)(x-b)-f=(x-c)(x-d)implies (x-c)(x-d)+f=(x-a)(x-b)$$



        Easy to see that this generalizes to higher degree.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Aug 2 at 12:57









        lulu

        34.7k14071




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