Proving roots to be real

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Given :



$$(a-b)x^2 + (c-a)x + (a-b) = 0 text where a,b,c in Q $$



How would I prove that the roots of the equation will be real






roots quadratics







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  • As pointed out by the other answer, it's not always possible. However, if you change the last $(a-b)$ to $(b-a)$, you'd have a much better starting point.
    – Arthur
    Sep 4 at 10:08










  • Use the discriminant.
    – Wuestenfux
    Sep 4 at 10:16














up vote
0
down vote

favorite












Given :



$$(a-b)x^2 + (c-a)x + (a-b) = 0 text where a,b,c in Q $$



How would I prove that the roots of the equation will be real






roots quadratics







share|cite|improve this question





















  • As pointed out by the other answer, it's not always possible. However, if you change the last $(a-b)$ to $(b-a)$, you'd have a much better starting point.
    – Arthur
    Sep 4 at 10:08










  • Use the discriminant.
    – Wuestenfux
    Sep 4 at 10:16












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Given :



$$(a-b)x^2 + (c-a)x + (a-b) = 0 text where a,b,c in Q $$



How would I prove that the roots of the equation will be real






roots quadratics







share|cite|improve this question













Given :



$$(a-b)x^2 + (c-a)x + (a-b) = 0 text where a,b,c in Q $$



How would I prove that the roots of the equation will be real






roots quadratics




roots quadratics






share|cite|improve this question













share|cite|improve this question











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share|cite|improve this question










asked Sep 4 at 10:05









Any3nymous user

101




101











  • As pointed out by the other answer, it's not always possible. However, if you change the last $(a-b)$ to $(b-a)$, you'd have a much better starting point.
    – Arthur
    Sep 4 at 10:08










  • Use the discriminant.
    – Wuestenfux
    Sep 4 at 10:16
















  • As pointed out by the other answer, it's not always possible. However, if you change the last $(a-b)$ to $(b-a)$, you'd have a much better starting point.
    – Arthur
    Sep 4 at 10:08










  • Use the discriminant.
    – Wuestenfux
    Sep 4 at 10:16















As pointed out by the other answer, it's not always possible. However, if you change the last $(a-b)$ to $(b-a)$, you'd have a much better starting point.
– Arthur
Sep 4 at 10:08




As pointed out by the other answer, it's not always possible. However, if you change the last $(a-b)$ to $(b-a)$, you'd have a much better starting point.
– Arthur
Sep 4 at 10:08












Use the discriminant.
– Wuestenfux
Sep 4 at 10:16




Use the discriminant.
– Wuestenfux
Sep 4 at 10:16










2 Answers
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You can't. If $b = 0$ and $a = c = 1$, then the roots of the equation are $pm i$.






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    down vote













    You have a quadratic equation of the form $ax^2 + bx + c$ being $a,b,c∈Q$ and $a = c$ in quadratic equations have two real solutions if $b^2 - 4ac gt 0$ (Which is the discriminant of the quadratic formula) that's why they are real solutions and not complex solutions. Now change the terms and you'll see.



    Note: You can get a real solution too if $b² - 4ac = 0$, but just one real solution.






    share|cite|improve this answer




















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      2 Answers
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      2 Answers
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      You can't. If $b = 0$ and $a = c = 1$, then the roots of the equation are $pm i$.






      share|cite|improve this answer
























        up vote
        4
        down vote













        You can't. If $b = 0$ and $a = c = 1$, then the roots of the equation are $pm i$.






        share|cite|improve this answer






















          up vote
          4
          down vote










          up vote
          4
          down vote









          You can't. If $b = 0$ and $a = c = 1$, then the roots of the equation are $pm i$.






          share|cite|improve this answer












          You can't. If $b = 0$ and $a = c = 1$, then the roots of the equation are $pm i$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Sep 4 at 10:07









          Mees de Vries

          14.3k12348




          14.3k12348




















              up vote
              0
              down vote













              You have a quadratic equation of the form $ax^2 + bx + c$ being $a,b,c∈Q$ and $a = c$ in quadratic equations have two real solutions if $b^2 - 4ac gt 0$ (Which is the discriminant of the quadratic formula) that's why they are real solutions and not complex solutions. Now change the terms and you'll see.



              Note: You can get a real solution too if $b² - 4ac = 0$, but just one real solution.






              share|cite|improve this answer
























                up vote
                0
                down vote













                You have a quadratic equation of the form $ax^2 + bx + c$ being $a,b,c∈Q$ and $a = c$ in quadratic equations have two real solutions if $b^2 - 4ac gt 0$ (Which is the discriminant of the quadratic formula) that's why they are real solutions and not complex solutions. Now change the terms and you'll see.



                Note: You can get a real solution too if $b² - 4ac = 0$, but just one real solution.






                share|cite|improve this answer






















                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  You have a quadratic equation of the form $ax^2 + bx + c$ being $a,b,c∈Q$ and $a = c$ in quadratic equations have two real solutions if $b^2 - 4ac gt 0$ (Which is the discriminant of the quadratic formula) that's why they are real solutions and not complex solutions. Now change the terms and you'll see.



                  Note: You can get a real solution too if $b² - 4ac = 0$, but just one real solution.






                  share|cite|improve this answer












                  You have a quadratic equation of the form $ax^2 + bx + c$ being $a,b,c∈Q$ and $a = c$ in quadratic equations have two real solutions if $b^2 - 4ac gt 0$ (Which is the discriminant of the quadratic formula) that's why they are real solutions and not complex solutions. Now change the terms and you'll see.



                  Note: You can get a real solution too if $b² - 4ac = 0$, but just one real solution.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Sep 4 at 11:02









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