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












2












$begingroup$


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











$endgroup$

















    2












    $begingroup$


    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











    $endgroup$















      2












      2








      2


      2



      $begingroup$


      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











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













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      share|cite|improve this question








      edited Nov 24 '18 at 5:33









      user21820

      38.9k543153




      38.9k543153










      asked Nov 23 '18 at 21:54









      guestguest

      4,232919




      4,232919






















          4 Answers
          4






          active

          oldest

          votes


















          10












          $begingroup$

          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











          $endgroup$









          • 1




            $begingroup$
            Well that makes my solution a little embarrassing.
            $endgroup$
            – guest
            Nov 23 '18 at 22:16






          • 5




            $begingroup$
            @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.
            $endgroup$
            – Martin R
            Nov 23 '18 at 22:51





















          6












          $begingroup$

          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









          $endgroup$













          • $begingroup$
            $b^2-4ac = (b+2c)^2 - 4(a+b+c)c > 0$ can be verified directly, without using determinants.
            $endgroup$
            – Martin R
            Nov 23 '18 at 22:11










          • $begingroup$
            @MartinR thanks! I am just showing how I arrived at the inequality :) Please post this as an answer so I will accept it.
            $endgroup$
            – guest
            Nov 23 '18 at 22:12





















          3












          $begingroup$

          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









          $endgroup$





















            2












            $begingroup$

            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.



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



            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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              active

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              10












              $begingroup$

              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











              $endgroup$









              • 1




                $begingroup$
                Well that makes my solution a little embarrassing.
                $endgroup$
                – guest
                Nov 23 '18 at 22:16






              • 5




                $begingroup$
                @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.
                $endgroup$
                – Martin R
                Nov 23 '18 at 22:51


















              10












              $begingroup$

              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











              $endgroup$









              • 1




                $begingroup$
                Well that makes my solution a little embarrassing.
                $endgroup$
                – guest
                Nov 23 '18 at 22:16






              • 5




                $begingroup$
                @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.
                $endgroup$
                – Martin R
                Nov 23 '18 at 22:51
















              10












              10








              10





              $begingroup$

              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











              $endgroup$



              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 '18 at 22:50

























              answered Nov 23 '18 at 22:15









              Martin RMartin R

              28.7k33356




              28.7k33356








              • 1




                $begingroup$
                Well that makes my solution a little embarrassing.
                $endgroup$
                – guest
                Nov 23 '18 at 22:16






              • 5




                $begingroup$
                @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.
                $endgroup$
                – Martin R
                Nov 23 '18 at 22:51
















              • 1




                $begingroup$
                Well that makes my solution a little embarrassing.
                $endgroup$
                – guest
                Nov 23 '18 at 22:16






              • 5




                $begingroup$
                @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.
                $endgroup$
                – Martin R
                Nov 23 '18 at 22:51










              1




              1




              $begingroup$
              Well that makes my solution a little embarrassing.
              $endgroup$
              – guest
              Nov 23 '18 at 22:16




              $begingroup$
              Well that makes my solution a little embarrassing.
              $endgroup$
              – guest
              Nov 23 '18 at 22:16




              5




              5




              $begingroup$
              @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.
              $endgroup$
              – Martin R
              Nov 23 '18 at 22:51






              $begingroup$
              @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.
              $endgroup$
              – Martin R
              Nov 23 '18 at 22:51













              6












              $begingroup$

              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









              $endgroup$













              • $begingroup$
                $b^2-4ac = (b+2c)^2 - 4(a+b+c)c > 0$ can be verified directly, without using determinants.
                $endgroup$
                – Martin R
                Nov 23 '18 at 22:11










              • $begingroup$
                @MartinR thanks! I am just showing how I arrived at the inequality :) Please post this as an answer so I will accept it.
                $endgroup$
                – guest
                Nov 23 '18 at 22:12


















              6












              $begingroup$

              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









              $endgroup$













              • $begingroup$
                $b^2-4ac = (b+2c)^2 - 4(a+b+c)c > 0$ can be verified directly, without using determinants.
                $endgroup$
                – Martin R
                Nov 23 '18 at 22:11










              • $begingroup$
                @MartinR thanks! I am just showing how I arrived at the inequality :) Please post this as an answer so I will accept it.
                $endgroup$
                – guest
                Nov 23 '18 at 22:12
















              6












              6








              6





              $begingroup$

              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









              $endgroup$



              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 '18 at 21:55









              guestguest

              4,232919




              4,232919












              • $begingroup$
                $b^2-4ac = (b+2c)^2 - 4(a+b+c)c > 0$ can be verified directly, without using determinants.
                $endgroup$
                – Martin R
                Nov 23 '18 at 22:11










              • $begingroup$
                @MartinR thanks! I am just showing how I arrived at the inequality :) Please post this as an answer so I will accept it.
                $endgroup$
                – guest
                Nov 23 '18 at 22:12




















              • $begingroup$
                $b^2-4ac = (b+2c)^2 - 4(a+b+c)c > 0$ can be verified directly, without using determinants.
                $endgroup$
                – Martin R
                Nov 23 '18 at 22:11










              • $begingroup$
                @MartinR thanks! I am just showing how I arrived at the inequality :) Please post this as an answer so I will accept it.
                $endgroup$
                – guest
                Nov 23 '18 at 22:12


















              $begingroup$
              $b^2-4ac = (b+2c)^2 - 4(a+b+c)c > 0$ can be verified directly, without using determinants.
              $endgroup$
              – Martin R
              Nov 23 '18 at 22:11




              $begingroup$
              $b^2-4ac = (b+2c)^2 - 4(a+b+c)c > 0$ can be verified directly, without using determinants.
              $endgroup$
              – Martin R
              Nov 23 '18 at 22:11












              $begingroup$
              @MartinR thanks! I am just showing how I arrived at the inequality :) Please post this as an answer so I will accept it.
              $endgroup$
              – guest
              Nov 23 '18 at 22:12






              $begingroup$
              @MartinR thanks! I am just showing how I arrived at the inequality :) Please post this as an answer so I will accept it.
              $endgroup$
              – guest
              Nov 23 '18 at 22:12













              3












              $begingroup$

              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









              $endgroup$


















                3












                $begingroup$

                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









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  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









                  $endgroup$



                  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












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 24 '18 at 0:58









                  obscuransobscurans

                  1,152311




                  1,152311























                      2












                      $begingroup$

                      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.



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



                      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











                      $endgroup$


















                        2












                        $begingroup$

                        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.



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



                        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











                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          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.



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



                          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











                          $endgroup$



                          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.



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



                          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 Dec 27 '18 at 2:10

























                          answered Nov 24 '18 at 2:53









                          Gregory NisbetGregory Nisbet

                          598312




                          598312






























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