Alternative proof that if $a,b,c in mathbb{R}$ and $(a+b+c)c0$?











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A problem from a Moscow Olympiad states : let $a,b,c$ be real numbers such that $(a+b+c)c<0$. Show that $b^2-4ac>0$. There is a well known proof of this applying the IVT on the polynomial $f(x) = ax^2+bx+c$. Is there any proof that does not involve any calculus, including IVT?



The question emerged from a reddit post
https://www.reddit.com/r/math/comments/9znumt/theorems_of_single_variable_calculus/










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    up vote
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    favorite
    2












    A problem from a Moscow Olympiad states : let $a,b,c$ be real numbers such that $(a+b+c)c<0$. Show that $b^2-4ac>0$. There is a well known proof of this applying the IVT on the polynomial $f(x) = ax^2+bx+c$. Is there any proof that does not involve any calculus, including IVT?



    The question emerged from a reddit post
    https://www.reddit.com/r/math/comments/9znumt/theorems_of_single_variable_calculus/










    share|cite|improve this question


























      up vote
      2
      down vote

      favorite
      2









      up vote
      2
      down vote

      favorite
      2






      2





      A problem from a Moscow Olympiad states : let $a,b,c$ be real numbers such that $(a+b+c)c<0$. Show that $b^2-4ac>0$. There is a well known proof of this applying the IVT on the polynomial $f(x) = ax^2+bx+c$. Is there any proof that does not involve any calculus, including IVT?



      The question emerged from a reddit post
      https://www.reddit.com/r/math/comments/9znumt/theorems_of_single_variable_calculus/










      share|cite|improve this question















      A problem from a Moscow Olympiad states : let $a,b,c$ be real numbers such that $(a+b+c)c<0$. Show that $b^2-4ac>0$. There is a well known proof of this applying the IVT on the polynomial $f(x) = ax^2+bx+c$. Is there any proof that does not involve any calculus, including IVT?



      The question emerged from a reddit post
      https://www.reddit.com/r/math/comments/9znumt/theorems_of_single_variable_calculus/







      inequality contest-math






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      edited Nov 24 at 5:33









      user21820

      38k541150




      38k541150










      asked Nov 23 at 21:54









      guest

      4,242919




      4,242919






















          4 Answers
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          up vote
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          down vote



          accepted










          We have
          $$
          (b^2 - 4ac) + 4(a+b+c)c = (b+2c)^2 ge 0 \
          implies b^2 - 4ac ge - 4(a+b+c)c > 0
          $$



          To give proper credit: The above approach was found after reading
          guest's answer
          and is just a simplification of that solution.






          share|cite|improve this answer



















          • 1




            Well that makes my solution a little embarrassing.
            – guest
            Nov 23 at 22:16






          • 4




            @guest: No, it isn't. It is a normal process that one finds a “complicated” solution first, and only then realizes how it can be simplified.
            – Martin R
            Nov 23 at 22:51




















          up vote
          7
          down vote













          Here is a proof that works for any ordered commutative ring:



          Let $a,b,c$ be rationals, reals or elements of any non-trivial ordered commutative ring. Assume that $(a+b+c)c<0$. We show that $b^2-4ac>0$.



          Solution:



          First, observe that the hypothesis implies $(a+b)c<-c ^2$ and that $cneq 0$.
          Then form the matrix
          $$M = left( begin{array}{cc}
          2c & b \
          b & 2a \
          end{array}right)$$

          whose determinant is precicely $4ac-b^2$, and the matrix
          $$ S = left( begin{array}{cc}
          1 & 0 \
          2 & 1 \
          end{array}right)$$

          whose determinant is $1$. Now from the multiplicativity of the determinant $det(SM) =det(S)det(M)=det(M)$. But
          $$det(SM) = det
          left( begin{array}{cc}
          2c & b \
          4c+b & 2(a+b) \
          end{array}right) = 4c(a+b) - b^2-4bc.$$



          But recall that $c(a+b)<-c^2$ so $$4c(a+b)-b^2-4bc < -4c^2-b^2 -4bc = -(b+2c)^2.$$
          The last element is always non-positive and the inequality is strict, so $det(M)=4ac-b^2<0$.






          share|cite|improve this answer





















          • $b^2-4ac = (b+2c)^2 - 4(a+b+c)c > 0$ can be verified directly, without using determinants.
            – Martin R
            Nov 23 at 22:11










          • @MartinR thanks! I am just showing how I arrived at the inequality :) Please post this as an answer so I will accept it.
            – guest
            Nov 23 at 22:12




















          up vote
          3
          down vote













          If all you know is completing the square:



          Suppose $b^2-4acleq0$. Then we know the quadratic $ax^2+bx+c$ has at most one real root.



          By completing the square we can rewrite $ax^2+bx+c=aleft(left(x+frac{b}{2a}right)^2+d^2right)$ for some real $d$.



          We have $c=frac{b^2}{4a}+ad^2$. Returning to the original expression, complete all the squares:



          $$begin{align}(a+b+c)c&=left(a+b+frac{b^2}{4a}+ad^2right)left(frac{b^2}{4a}+ad^2right)\
          &=left(a+b+frac{b^2}{4a}right)frac{b^2}{4a}+left(ad^2frac{b^2}{4a}+left(a+b+frac{b^2}{4a}right)ad^2right)+left(ad^2right)^2\
          &=frac{b^2}{4}left(1+frac{b}{a}+frac{b^2}{4a^2}right)+d^2left(a^2+ab+frac{b^2}{2}right)+left(ad^2right)^2\
          &=frac{b^2}{4}left(1+frac{b}{2a}right)^2+d^2left(a+frac{b}{2}right)^2+left(ad^2right)^2\
          &geq0text{.}
          end{align}$$

          Thus by contrapositive, $(a+b+c)c<0,Rightarrow, b^2-4ac>0$.






          share|cite|improve this answer








          New contributor




          obscurans is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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            up vote
            2
            down vote













            Here's a silly proof deriving a contradiction from assuming that $(a + b + c)c lt 0 $ and $b^2-4ac le 0$ are simultaneously true.



            1) By assumption



            $$ (a + b + c) c lt 0 $$



            2) Negation of goal: $ b^2 - 4ac gt 0$



            $$ b^2 - 4ac le 0 $$



            3) From 2 by addition



            $$ b^2 le 4ac $$



            4) expand 1



            $$ ac + bc + c^2 < 0 $$



            5) From 4 by addition



            $$ ac < -bc - c^2 $$



            6) From 5 by multiplication



            $$ 4ac < -4bc -4c^2 $$



            7) transitivity $x le y lt z$ implies $x lt z$



            $$ b^2 lt -4bc - 4c^2 $$



            8) from 7 by addition.



            $$ b^2 + 4bc + 4c^2 gt 0 $$



            9) (7) is a perfect square



            $$ (b + 2c)^2 < 0 $$



            10) By assumption, $a, b, c in mathbb{R}$, so the square of $(b + 2c)$ cannot be negative.



            $$ bot $$



            Therefore, $b^2 - 4ac gt 0$ .






            share|cite|improve this answer























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              4 Answers
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              active

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              4 Answers
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              active

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              up vote
              10
              down vote



              accepted










              We have
              $$
              (b^2 - 4ac) + 4(a+b+c)c = (b+2c)^2 ge 0 \
              implies b^2 - 4ac ge - 4(a+b+c)c > 0
              $$



              To give proper credit: The above approach was found after reading
              guest's answer
              and is just a simplification of that solution.






              share|cite|improve this answer



















              • 1




                Well that makes my solution a little embarrassing.
                – guest
                Nov 23 at 22:16






              • 4




                @guest: No, it isn't. It is a normal process that one finds a “complicated” solution first, and only then realizes how it can be simplified.
                – Martin R
                Nov 23 at 22:51

















              up vote
              10
              down vote



              accepted










              We have
              $$
              (b^2 - 4ac) + 4(a+b+c)c = (b+2c)^2 ge 0 \
              implies b^2 - 4ac ge - 4(a+b+c)c > 0
              $$



              To give proper credit: The above approach was found after reading
              guest's answer
              and is just a simplification of that solution.






              share|cite|improve this answer



















              • 1




                Well that makes my solution a little embarrassing.
                – guest
                Nov 23 at 22:16






              • 4




                @guest: No, it isn't. It is a normal process that one finds a “complicated” solution first, and only then realizes how it can be simplified.
                – Martin R
                Nov 23 at 22:51















              up vote
              10
              down vote



              accepted







              up vote
              10
              down vote



              accepted






              We have
              $$
              (b^2 - 4ac) + 4(a+b+c)c = (b+2c)^2 ge 0 \
              implies b^2 - 4ac ge - 4(a+b+c)c > 0
              $$



              To give proper credit: The above approach was found after reading
              guest's answer
              and is just a simplification of that solution.






              share|cite|improve this answer














              We have
              $$
              (b^2 - 4ac) + 4(a+b+c)c = (b+2c)^2 ge 0 \
              implies b^2 - 4ac ge - 4(a+b+c)c > 0
              $$



              To give proper credit: The above approach was found after reading
              guest's answer
              and is just a simplification of that solution.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Nov 23 at 22:50

























              answered Nov 23 at 22:15









              Martin R

              25.9k32946




              25.9k32946








              • 1




                Well that makes my solution a little embarrassing.
                – guest
                Nov 23 at 22:16






              • 4




                @guest: No, it isn't. It is a normal process that one finds a “complicated” solution first, and only then realizes how it can be simplified.
                – Martin R
                Nov 23 at 22:51
















              • 1




                Well that makes my solution a little embarrassing.
                – guest
                Nov 23 at 22:16






              • 4




                @guest: No, it isn't. It is a normal process that one finds a “complicated” solution first, and only then realizes how it can be simplified.
                – Martin R
                Nov 23 at 22:51










              1




              1




              Well that makes my solution a little embarrassing.
              – guest
              Nov 23 at 22:16




              Well that makes my solution a little embarrassing.
              – guest
              Nov 23 at 22:16




              4




              4




              @guest: No, it isn't. It is a normal process that one finds a “complicated” solution first, and only then realizes how it can be simplified.
              – Martin R
              Nov 23 at 22:51






              @guest: No, it isn't. It is a normal process that one finds a “complicated” solution first, and only then realizes how it can be simplified.
              – Martin R
              Nov 23 at 22:51












              up vote
              7
              down vote













              Here is a proof that works for any ordered commutative ring:



              Let $a,b,c$ be rationals, reals or elements of any non-trivial ordered commutative ring. Assume that $(a+b+c)c<0$. We show that $b^2-4ac>0$.



              Solution:



              First, observe that the hypothesis implies $(a+b)c<-c ^2$ and that $cneq 0$.
              Then form the matrix
              $$M = left( begin{array}{cc}
              2c & b \
              b & 2a \
              end{array}right)$$

              whose determinant is precicely $4ac-b^2$, and the matrix
              $$ S = left( begin{array}{cc}
              1 & 0 \
              2 & 1 \
              end{array}right)$$

              whose determinant is $1$. Now from the multiplicativity of the determinant $det(SM) =det(S)det(M)=det(M)$. But
              $$det(SM) = det
              left( begin{array}{cc}
              2c & b \
              4c+b & 2(a+b) \
              end{array}right) = 4c(a+b) - b^2-4bc.$$



              But recall that $c(a+b)<-c^2$ so $$4c(a+b)-b^2-4bc < -4c^2-b^2 -4bc = -(b+2c)^2.$$
              The last element is always non-positive and the inequality is strict, so $det(M)=4ac-b^2<0$.






              share|cite|improve this answer





















              • $b^2-4ac = (b+2c)^2 - 4(a+b+c)c > 0$ can be verified directly, without using determinants.
                – Martin R
                Nov 23 at 22:11










              • @MartinR thanks! I am just showing how I arrived at the inequality :) Please post this as an answer so I will accept it.
                – guest
                Nov 23 at 22:12

















              up vote
              7
              down vote













              Here is a proof that works for any ordered commutative ring:



              Let $a,b,c$ be rationals, reals or elements of any non-trivial ordered commutative ring. Assume that $(a+b+c)c<0$. We show that $b^2-4ac>0$.



              Solution:



              First, observe that the hypothesis implies $(a+b)c<-c ^2$ and that $cneq 0$.
              Then form the matrix
              $$M = left( begin{array}{cc}
              2c & b \
              b & 2a \
              end{array}right)$$

              whose determinant is precicely $4ac-b^2$, and the matrix
              $$ S = left( begin{array}{cc}
              1 & 0 \
              2 & 1 \
              end{array}right)$$

              whose determinant is $1$. Now from the multiplicativity of the determinant $det(SM) =det(S)det(M)=det(M)$. But
              $$det(SM) = det
              left( begin{array}{cc}
              2c & b \
              4c+b & 2(a+b) \
              end{array}right) = 4c(a+b) - b^2-4bc.$$



              But recall that $c(a+b)<-c^2$ so $$4c(a+b)-b^2-4bc < -4c^2-b^2 -4bc = -(b+2c)^2.$$
              The last element is always non-positive and the inequality is strict, so $det(M)=4ac-b^2<0$.






              share|cite|improve this answer





















              • $b^2-4ac = (b+2c)^2 - 4(a+b+c)c > 0$ can be verified directly, without using determinants.
                – Martin R
                Nov 23 at 22:11










              • @MartinR thanks! I am just showing how I arrived at the inequality :) Please post this as an answer so I will accept it.
                – guest
                Nov 23 at 22:12















              up vote
              7
              down vote










              up vote
              7
              down vote









              Here is a proof that works for any ordered commutative ring:



              Let $a,b,c$ be rationals, reals or elements of any non-trivial ordered commutative ring. Assume that $(a+b+c)c<0$. We show that $b^2-4ac>0$.



              Solution:



              First, observe that the hypothesis implies $(a+b)c<-c ^2$ and that $cneq 0$.
              Then form the matrix
              $$M = left( begin{array}{cc}
              2c & b \
              b & 2a \
              end{array}right)$$

              whose determinant is precicely $4ac-b^2$, and the matrix
              $$ S = left( begin{array}{cc}
              1 & 0 \
              2 & 1 \
              end{array}right)$$

              whose determinant is $1$. Now from the multiplicativity of the determinant $det(SM) =det(S)det(M)=det(M)$. But
              $$det(SM) = det
              left( begin{array}{cc}
              2c & b \
              4c+b & 2(a+b) \
              end{array}right) = 4c(a+b) - b^2-4bc.$$



              But recall that $c(a+b)<-c^2$ so $$4c(a+b)-b^2-4bc < -4c^2-b^2 -4bc = -(b+2c)^2.$$
              The last element is always non-positive and the inequality is strict, so $det(M)=4ac-b^2<0$.






              share|cite|improve this answer












              Here is a proof that works for any ordered commutative ring:



              Let $a,b,c$ be rationals, reals or elements of any non-trivial ordered commutative ring. Assume that $(a+b+c)c<0$. We show that $b^2-4ac>0$.



              Solution:



              First, observe that the hypothesis implies $(a+b)c<-c ^2$ and that $cneq 0$.
              Then form the matrix
              $$M = left( begin{array}{cc}
              2c & b \
              b & 2a \
              end{array}right)$$

              whose determinant is precicely $4ac-b^2$, and the matrix
              $$ S = left( begin{array}{cc}
              1 & 0 \
              2 & 1 \
              end{array}right)$$

              whose determinant is $1$. Now from the multiplicativity of the determinant $det(SM) =det(S)det(M)=det(M)$. But
              $$det(SM) = det
              left( begin{array}{cc}
              2c & b \
              4c+b & 2(a+b) \
              end{array}right) = 4c(a+b) - b^2-4bc.$$



              But recall that $c(a+b)<-c^2$ so $$4c(a+b)-b^2-4bc < -4c^2-b^2 -4bc = -(b+2c)^2.$$
              The last element is always non-positive and the inequality is strict, so $det(M)=4ac-b^2<0$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 23 at 21:55









              guest

              4,242919




              4,242919












              • $b^2-4ac = (b+2c)^2 - 4(a+b+c)c > 0$ can be verified directly, without using determinants.
                – Martin R
                Nov 23 at 22:11










              • @MartinR thanks! I am just showing how I arrived at the inequality :) Please post this as an answer so I will accept it.
                – guest
                Nov 23 at 22:12




















              • $b^2-4ac = (b+2c)^2 - 4(a+b+c)c > 0$ can be verified directly, without using determinants.
                – Martin R
                Nov 23 at 22:11










              • @MartinR thanks! I am just showing how I arrived at the inequality :) Please post this as an answer so I will accept it.
                – guest
                Nov 23 at 22:12


















              $b^2-4ac = (b+2c)^2 - 4(a+b+c)c > 0$ can be verified directly, without using determinants.
              – Martin R
              Nov 23 at 22:11




              $b^2-4ac = (b+2c)^2 - 4(a+b+c)c > 0$ can be verified directly, without using determinants.
              – Martin R
              Nov 23 at 22:11












              @MartinR thanks! I am just showing how I arrived at the inequality :) Please post this as an answer so I will accept it.
              – guest
              Nov 23 at 22:12






              @MartinR thanks! I am just showing how I arrived at the inequality :) Please post this as an answer so I will accept it.
              – guest
              Nov 23 at 22:12












              up vote
              3
              down vote













              If all you know is completing the square:



              Suppose $b^2-4acleq0$. Then we know the quadratic $ax^2+bx+c$ has at most one real root.



              By completing the square we can rewrite $ax^2+bx+c=aleft(left(x+frac{b}{2a}right)^2+d^2right)$ for some real $d$.



              We have $c=frac{b^2}{4a}+ad^2$. Returning to the original expression, complete all the squares:



              $$begin{align}(a+b+c)c&=left(a+b+frac{b^2}{4a}+ad^2right)left(frac{b^2}{4a}+ad^2right)\
              &=left(a+b+frac{b^2}{4a}right)frac{b^2}{4a}+left(ad^2frac{b^2}{4a}+left(a+b+frac{b^2}{4a}right)ad^2right)+left(ad^2right)^2\
              &=frac{b^2}{4}left(1+frac{b}{a}+frac{b^2}{4a^2}right)+d^2left(a^2+ab+frac{b^2}{2}right)+left(ad^2right)^2\
              &=frac{b^2}{4}left(1+frac{b}{2a}right)^2+d^2left(a+frac{b}{2}right)^2+left(ad^2right)^2\
              &geq0text{.}
              end{align}$$

              Thus by contrapositive, $(a+b+c)c<0,Rightarrow, b^2-4ac>0$.






              share|cite|improve this answer








              New contributor




              obscurans is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.






















                up vote
                3
                down vote













                If all you know is completing the square:



                Suppose $b^2-4acleq0$. Then we know the quadratic $ax^2+bx+c$ has at most one real root.



                By completing the square we can rewrite $ax^2+bx+c=aleft(left(x+frac{b}{2a}right)^2+d^2right)$ for some real $d$.



                We have $c=frac{b^2}{4a}+ad^2$. Returning to the original expression, complete all the squares:



                $$begin{align}(a+b+c)c&=left(a+b+frac{b^2}{4a}+ad^2right)left(frac{b^2}{4a}+ad^2right)\
                &=left(a+b+frac{b^2}{4a}right)frac{b^2}{4a}+left(ad^2frac{b^2}{4a}+left(a+b+frac{b^2}{4a}right)ad^2right)+left(ad^2right)^2\
                &=frac{b^2}{4}left(1+frac{b}{a}+frac{b^2}{4a^2}right)+d^2left(a^2+ab+frac{b^2}{2}right)+left(ad^2right)^2\
                &=frac{b^2}{4}left(1+frac{b}{2a}right)^2+d^2left(a+frac{b}{2}right)^2+left(ad^2right)^2\
                &geq0text{.}
                end{align}$$

                Thus by contrapositive, $(a+b+c)c<0,Rightarrow, b^2-4ac>0$.






                share|cite|improve this answer








                New contributor




                obscurans is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.




















                  up vote
                  3
                  down vote










                  up vote
                  3
                  down vote









                  If all you know is completing the square:



                  Suppose $b^2-4acleq0$. Then we know the quadratic $ax^2+bx+c$ has at most one real root.



                  By completing the square we can rewrite $ax^2+bx+c=aleft(left(x+frac{b}{2a}right)^2+d^2right)$ for some real $d$.



                  We have $c=frac{b^2}{4a}+ad^2$. Returning to the original expression, complete all the squares:



                  $$begin{align}(a+b+c)c&=left(a+b+frac{b^2}{4a}+ad^2right)left(frac{b^2}{4a}+ad^2right)\
                  &=left(a+b+frac{b^2}{4a}right)frac{b^2}{4a}+left(ad^2frac{b^2}{4a}+left(a+b+frac{b^2}{4a}right)ad^2right)+left(ad^2right)^2\
                  &=frac{b^2}{4}left(1+frac{b}{a}+frac{b^2}{4a^2}right)+d^2left(a^2+ab+frac{b^2}{2}right)+left(ad^2right)^2\
                  &=frac{b^2}{4}left(1+frac{b}{2a}right)^2+d^2left(a+frac{b}{2}right)^2+left(ad^2right)^2\
                  &geq0text{.}
                  end{align}$$

                  Thus by contrapositive, $(a+b+c)c<0,Rightarrow, b^2-4ac>0$.






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                  If all you know is completing the square:



                  Suppose $b^2-4acleq0$. Then we know the quadratic $ax^2+bx+c$ has at most one real root.



                  By completing the square we can rewrite $ax^2+bx+c=aleft(left(x+frac{b}{2a}right)^2+d^2right)$ for some real $d$.



                  We have $c=frac{b^2}{4a}+ad^2$. Returning to the original expression, complete all the squares:



                  $$begin{align}(a+b+c)c&=left(a+b+frac{b^2}{4a}+ad^2right)left(frac{b^2}{4a}+ad^2right)\
                  &=left(a+b+frac{b^2}{4a}right)frac{b^2}{4a}+left(ad^2frac{b^2}{4a}+left(a+b+frac{b^2}{4a}right)ad^2right)+left(ad^2right)^2\
                  &=frac{b^2}{4}left(1+frac{b}{a}+frac{b^2}{4a^2}right)+d^2left(a^2+ab+frac{b^2}{2}right)+left(ad^2right)^2\
                  &=frac{b^2}{4}left(1+frac{b}{2a}right)^2+d^2left(a+frac{b}{2}right)^2+left(ad^2right)^2\
                  &geq0text{.}
                  end{align}$$

                  Thus by contrapositive, $(a+b+c)c<0,Rightarrow, b^2-4ac>0$.







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                  share|cite|improve this answer



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                  answered Nov 24 at 0:58









                  obscurans

                  664




                  664




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                      up vote
                      2
                      down vote













                      Here's a silly proof deriving a contradiction from assuming that $(a + b + c)c lt 0 $ and $b^2-4ac le 0$ are simultaneously true.



                      1) By assumption



                      $$ (a + b + c) c lt 0 $$



                      2) Negation of goal: $ b^2 - 4ac gt 0$



                      $$ b^2 - 4ac le 0 $$



                      3) From 2 by addition



                      $$ b^2 le 4ac $$



                      4) expand 1



                      $$ ac + bc + c^2 < 0 $$



                      5) From 4 by addition



                      $$ ac < -bc - c^2 $$



                      6) From 5 by multiplication



                      $$ 4ac < -4bc -4c^2 $$



                      7) transitivity $x le y lt z$ implies $x lt z$



                      $$ b^2 lt -4bc - 4c^2 $$



                      8) from 7 by addition.



                      $$ b^2 + 4bc + 4c^2 gt 0 $$



                      9) (7) is a perfect square



                      $$ (b + 2c)^2 < 0 $$



                      10) By assumption, $a, b, c in mathbb{R}$, so the square of $(b + 2c)$ cannot be negative.



                      $$ bot $$



                      Therefore, $b^2 - 4ac gt 0$ .






                      share|cite|improve this answer



























                        up vote
                        2
                        down vote













                        Here's a silly proof deriving a contradiction from assuming that $(a + b + c)c lt 0 $ and $b^2-4ac le 0$ are simultaneously true.



                        1) By assumption



                        $$ (a + b + c) c lt 0 $$



                        2) Negation of goal: $ b^2 - 4ac gt 0$



                        $$ b^2 - 4ac le 0 $$



                        3) From 2 by addition



                        $$ b^2 le 4ac $$



                        4) expand 1



                        $$ ac + bc + c^2 < 0 $$



                        5) From 4 by addition



                        $$ ac < -bc - c^2 $$



                        6) From 5 by multiplication



                        $$ 4ac < -4bc -4c^2 $$



                        7) transitivity $x le y lt z$ implies $x lt z$



                        $$ b^2 lt -4bc - 4c^2 $$



                        8) from 7 by addition.



                        $$ b^2 + 4bc + 4c^2 gt 0 $$



                        9) (7) is a perfect square



                        $$ (b + 2c)^2 < 0 $$



                        10) By assumption, $a, b, c in mathbb{R}$, so the square of $(b + 2c)$ cannot be negative.



                        $$ bot $$



                        Therefore, $b^2 - 4ac gt 0$ .






                        share|cite|improve this answer

























                          up vote
                          2
                          down vote










                          up vote
                          2
                          down vote









                          Here's a silly proof deriving a contradiction from assuming that $(a + b + c)c lt 0 $ and $b^2-4ac le 0$ are simultaneously true.



                          1) By assumption



                          $$ (a + b + c) c lt 0 $$



                          2) Negation of goal: $ b^2 - 4ac gt 0$



                          $$ b^2 - 4ac le 0 $$



                          3) From 2 by addition



                          $$ b^2 le 4ac $$



                          4) expand 1



                          $$ ac + bc + c^2 < 0 $$



                          5) From 4 by addition



                          $$ ac < -bc - c^2 $$



                          6) From 5 by multiplication



                          $$ 4ac < -4bc -4c^2 $$



                          7) transitivity $x le y lt z$ implies $x lt z$



                          $$ b^2 lt -4bc - 4c^2 $$



                          8) from 7 by addition.



                          $$ b^2 + 4bc + 4c^2 gt 0 $$



                          9) (7) is a perfect square



                          $$ (b + 2c)^2 < 0 $$



                          10) By assumption, $a, b, c in mathbb{R}$, so the square of $(b + 2c)$ cannot be negative.



                          $$ bot $$



                          Therefore, $b^2 - 4ac gt 0$ .






                          share|cite|improve this answer














                          Here's a silly proof deriving a contradiction from assuming that $(a + b + c)c lt 0 $ and $b^2-4ac le 0$ are simultaneously true.



                          1) By assumption



                          $$ (a + b + c) c lt 0 $$



                          2) Negation of goal: $ b^2 - 4ac gt 0$



                          $$ b^2 - 4ac le 0 $$



                          3) From 2 by addition



                          $$ b^2 le 4ac $$



                          4) expand 1



                          $$ ac + bc + c^2 < 0 $$



                          5) From 4 by addition



                          $$ ac < -bc - c^2 $$



                          6) From 5 by multiplication



                          $$ 4ac < -4bc -4c^2 $$



                          7) transitivity $x le y lt z$ implies $x lt z$



                          $$ b^2 lt -4bc - 4c^2 $$



                          8) from 7 by addition.



                          $$ b^2 + 4bc + 4c^2 gt 0 $$



                          9) (7) is a perfect square



                          $$ (b + 2c)^2 < 0 $$



                          10) By assumption, $a, b, c in mathbb{R}$, so the square of $(b + 2c)$ cannot be negative.



                          $$ bot $$



                          Therefore, $b^2 - 4ac gt 0$ .







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Nov 24 at 4:52

























                          answered Nov 24 at 2:53









                          Gregory Nisbet

                          368211




                          368211






























                               

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