Solve quadratic equation by completing square












0














We have a equation: $2x^2+7x+3$

I tried to find the vertex of the parabola by this formula: $a(x-h)^2+k$

but I could not get it.

I got this: but it is not right.
$2(x^2+ frac{7}{2}x+frac{49}{4})+3-frac{49}{2}$

What's the problem? Is it possible to solve it by completing square?






quadratics







share|cite|improve this question
























  • Why do you want to find the vertex by strictly completing the square ?
    – Rebellos
    Nov 24 at 11:13












  • Because my book is given this formula.
    – Ehsan Zehtabchi
    Nov 24 at 11:16










  • The formula makes no sense. You don't need a negative sign there.
    – Rebellos
    Nov 24 at 11:18
















0














We have a equation: $2x^2+7x+3$

I tried to find the vertex of the parabola by this formula: $a(x-h)^2+k$

but I could not get it.

I got this: but it is not right.
$2(x^2+ frac{7}{2}x+frac{49}{4})+3-frac{49}{2}$

What's the problem? Is it possible to solve it by completing square?






quadratics







share|cite|improve this question
























  • Why do you want to find the vertex by strictly completing the square ?
    – Rebellos
    Nov 24 at 11:13












  • Because my book is given this formula.
    – Ehsan Zehtabchi
    Nov 24 at 11:16










  • The formula makes no sense. You don't need a negative sign there.
    – Rebellos
    Nov 24 at 11:18














0












0








0







We have a equation: $2x^2+7x+3$

I tried to find the vertex of the parabola by this formula: $a(x-h)^2+k$

but I could not get it.

I got this: but it is not right.
$2(x^2+ frac{7}{2}x+frac{49}{4})+3-frac{49}{2}$

What's the problem? Is it possible to solve it by completing square?






quadratics







share|cite|improve this question















We have a equation: $2x^2+7x+3$

I tried to find the vertex of the parabola by this formula: $a(x-h)^2+k$

but I could not get it.

I got this: but it is not right.
$2(x^2+ frac{7}{2}x+frac{49}{4})+3-frac{49}{2}$

What's the problem? Is it possible to solve it by completing square?






quadratics




quadratics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 24 at 14:59









Dave L. Renfro

24.3k33979




24.3k33979










asked Nov 24 at 11:09









Ehsan Zehtabchi

263




263












  • Why do you want to find the vertex by strictly completing the square ?
    – Rebellos
    Nov 24 at 11:13












  • Because my book is given this formula.
    – Ehsan Zehtabchi
    Nov 24 at 11:16










  • The formula makes no sense. You don't need a negative sign there.
    – Rebellos
    Nov 24 at 11:18


















  • Why do you want to find the vertex by strictly completing the square ?
    – Rebellos
    Nov 24 at 11:13












  • Because my book is given this formula.
    – Ehsan Zehtabchi
    Nov 24 at 11:16










  • The formula makes no sense. You don't need a negative sign there.
    – Rebellos
    Nov 24 at 11:18
















Why do you want to find the vertex by strictly completing the square ?
– Rebellos
Nov 24 at 11:13






Why do you want to find the vertex by strictly completing the square ?
– Rebellos
Nov 24 at 11:13














Because my book is given this formula.
– Ehsan Zehtabchi
Nov 24 at 11:16




Because my book is given this formula.
– Ehsan Zehtabchi
Nov 24 at 11:16












The formula makes no sense. You don't need a negative sign there.
– Rebellos
Nov 24 at 11:18




The formula makes no sense. You don't need a negative sign there.
– Rebellos
Nov 24 at 11:18










3 Answers
3






active

oldest

votes


















3














Completing the square, it is :



$$2x^2 + 7x + 3 = 2bigg(x^2+ frac{7}{2}x + frac{49}{16}bigg) - frac{25}{8} = 2bigg(x + frac{7}{4}bigg)^2 - frac{25}{8}$$



Thus, the vertex can be found at the point where the "completed square term" equals zero :



$$x+frac{7}{4} = 0 Leftrightarrow x = -frac{7}{4}$$



This is truly the vertex point, as it can be cross-validated by the fact that a vertex of a parabola can be found at :



$$x_v = -frac{b}{2a} Rightarrow x_v = -frac{7}{2cdot 2}$$



Now, the other coordinate of the vertex is the remaining constant term on the right, thus $-25/8$.



This means that the given vertex coordinates, are :



$$(h,k) = bigg(-frac{7}{4},-frac{25}{8}bigg)$$






share|cite|improve this answer





























    2














    It's possible.



    begin{align*}
    2x^2 + 7x + 3
    & = 2left(x^2 + frac{7}{2} x right) + 3
    = 2left(left(x + frac{7}{4}right)^2 - left(frac{7}{4}right)^2 right) + 3 \
    & = 2 left(x + frac{7}{4}right)^2 - frac{49}{8} + 3
    = 2 left(x + frac{7}{4}right)^2 - frac{25}{8}
    end{align*}
    So we have the vertex at $(h,k) = left(- frac{7}{4}, - frac{25}{8}right)$.






    share|cite|improve this answer





























      2














      In completing the square, you want to follow a plan like below.



      $$ax^2+bx+c = 0$$



      $$x^2+frac{b}{a}x+frac{c}{a} = 0 implies x^2+frac{b}{a}x = -frac{c}{a}$$



      You want to reach a quadratic trinomial on the LHS so you can solve it. This is done by adding $big(frac{b}{2a}big)^2$ to both sides.



      $$x^2+frac{b}{a}x+bigg(frac{b}{2a}bigg)^2 = -frac{c}{a}+frac{b^2}{4a^2}$$



      $$bigg(x+frac{b}{2a}bigg)^2 = frac{b^2-4ac}{4a^2}$$



      After which you can take the square root of both sides and solve for $x$. Can you use this general plan to solve the equation you’ve given?






      share|cite|improve this answer





















        Your Answer





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        3 Answers
        3






        active

        oldest

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        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3














        Completing the square, it is :



        $$2x^2 + 7x + 3 = 2bigg(x^2+ frac{7}{2}x + frac{49}{16}bigg) - frac{25}{8} = 2bigg(x + frac{7}{4}bigg)^2 - frac{25}{8}$$



        Thus, the vertex can be found at the point where the "completed square term" equals zero :



        $$x+frac{7}{4} = 0 Leftrightarrow x = -frac{7}{4}$$



        This is truly the vertex point, as it can be cross-validated by the fact that a vertex of a parabola can be found at :



        $$x_v = -frac{b}{2a} Rightarrow x_v = -frac{7}{2cdot 2}$$



        Now, the other coordinate of the vertex is the remaining constant term on the right, thus $-25/8$.



        This means that the given vertex coordinates, are :



        $$(h,k) = bigg(-frac{7}{4},-frac{25}{8}bigg)$$






        share|cite|improve this answer


























          3














          Completing the square, it is :



          $$2x^2 + 7x + 3 = 2bigg(x^2+ frac{7}{2}x + frac{49}{16}bigg) - frac{25}{8} = 2bigg(x + frac{7}{4}bigg)^2 - frac{25}{8}$$



          Thus, the vertex can be found at the point where the "completed square term" equals zero :



          $$x+frac{7}{4} = 0 Leftrightarrow x = -frac{7}{4}$$



          This is truly the vertex point, as it can be cross-validated by the fact that a vertex of a parabola can be found at :



          $$x_v = -frac{b}{2a} Rightarrow x_v = -frac{7}{2cdot 2}$$



          Now, the other coordinate of the vertex is the remaining constant term on the right, thus $-25/8$.



          This means that the given vertex coordinates, are :



          $$(h,k) = bigg(-frac{7}{4},-frac{25}{8}bigg)$$






          share|cite|improve this answer
























            3












            3








            3






            Completing the square, it is :



            $$2x^2 + 7x + 3 = 2bigg(x^2+ frac{7}{2}x + frac{49}{16}bigg) - frac{25}{8} = 2bigg(x + frac{7}{4}bigg)^2 - frac{25}{8}$$



            Thus, the vertex can be found at the point where the "completed square term" equals zero :



            $$x+frac{7}{4} = 0 Leftrightarrow x = -frac{7}{4}$$



            This is truly the vertex point, as it can be cross-validated by the fact that a vertex of a parabola can be found at :



            $$x_v = -frac{b}{2a} Rightarrow x_v = -frac{7}{2cdot 2}$$



            Now, the other coordinate of the vertex is the remaining constant term on the right, thus $-25/8$.



            This means that the given vertex coordinates, are :



            $$(h,k) = bigg(-frac{7}{4},-frac{25}{8}bigg)$$






            share|cite|improve this answer












            Completing the square, it is :



            $$2x^2 + 7x + 3 = 2bigg(x^2+ frac{7}{2}x + frac{49}{16}bigg) - frac{25}{8} = 2bigg(x + frac{7}{4}bigg)^2 - frac{25}{8}$$



            Thus, the vertex can be found at the point where the "completed square term" equals zero :



            $$x+frac{7}{4} = 0 Leftrightarrow x = -frac{7}{4}$$



            This is truly the vertex point, as it can be cross-validated by the fact that a vertex of a parabola can be found at :



            $$x_v = -frac{b}{2a} Rightarrow x_v = -frac{7}{2cdot 2}$$



            Now, the other coordinate of the vertex is the remaining constant term on the right, thus $-25/8$.



            This means that the given vertex coordinates, are :



            $$(h,k) = bigg(-frac{7}{4},-frac{25}{8}bigg)$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 24 at 11:21









            Rebellos

            14.2k31244




            14.2k31244























                2














                It's possible.



                begin{align*}
                2x^2 + 7x + 3
                & = 2left(x^2 + frac{7}{2} x right) + 3
                = 2left(left(x + frac{7}{4}right)^2 - left(frac{7}{4}right)^2 right) + 3 \
                & = 2 left(x + frac{7}{4}right)^2 - frac{49}{8} + 3
                = 2 left(x + frac{7}{4}right)^2 - frac{25}{8}
                end{align*}
                So we have the vertex at $(h,k) = left(- frac{7}{4}, - frac{25}{8}right)$.






                share|cite|improve this answer


























                  2














                  It's possible.



                  begin{align*}
                  2x^2 + 7x + 3
                  & = 2left(x^2 + frac{7}{2} x right) + 3
                  = 2left(left(x + frac{7}{4}right)^2 - left(frac{7}{4}right)^2 right) + 3 \
                  & = 2 left(x + frac{7}{4}right)^2 - frac{49}{8} + 3
                  = 2 left(x + frac{7}{4}right)^2 - frac{25}{8}
                  end{align*}
                  So we have the vertex at $(h,k) = left(- frac{7}{4}, - frac{25}{8}right)$.






                  share|cite|improve this answer
























                    2












                    2








                    2






                    It's possible.



                    begin{align*}
                    2x^2 + 7x + 3
                    & = 2left(x^2 + frac{7}{2} x right) + 3
                    = 2left(left(x + frac{7}{4}right)^2 - left(frac{7}{4}right)^2 right) + 3 \
                    & = 2 left(x + frac{7}{4}right)^2 - frac{49}{8} + 3
                    = 2 left(x + frac{7}{4}right)^2 - frac{25}{8}
                    end{align*}
                    So we have the vertex at $(h,k) = left(- frac{7}{4}, - frac{25}{8}right)$.






                    share|cite|improve this answer












                    It's possible.



                    begin{align*}
                    2x^2 + 7x + 3
                    & = 2left(x^2 + frac{7}{2} x right) + 3
                    = 2left(left(x + frac{7}{4}right)^2 - left(frac{7}{4}right)^2 right) + 3 \
                    & = 2 left(x + frac{7}{4}right)^2 - frac{49}{8} + 3
                    = 2 left(x + frac{7}{4}right)^2 - frac{25}{8}
                    end{align*}
                    So we have the vertex at $(h,k) = left(- frac{7}{4}, - frac{25}{8}right)$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 24 at 11:19









                    Viktor Glombik

                    544424




                    544424























                        2














                        In completing the square, you want to follow a plan like below.



                        $$ax^2+bx+c = 0$$



                        $$x^2+frac{b}{a}x+frac{c}{a} = 0 implies x^2+frac{b}{a}x = -frac{c}{a}$$



                        You want to reach a quadratic trinomial on the LHS so you can solve it. This is done by adding $big(frac{b}{2a}big)^2$ to both sides.



                        $$x^2+frac{b}{a}x+bigg(frac{b}{2a}bigg)^2 = -frac{c}{a}+frac{b^2}{4a^2}$$



                        $$bigg(x+frac{b}{2a}bigg)^2 = frac{b^2-4ac}{4a^2}$$



                        After which you can take the square root of both sides and solve for $x$. Can you use this general plan to solve the equation you’ve given?






                        share|cite|improve this answer


























                          2














                          In completing the square, you want to follow a plan like below.



                          $$ax^2+bx+c = 0$$



                          $$x^2+frac{b}{a}x+frac{c}{a} = 0 implies x^2+frac{b}{a}x = -frac{c}{a}$$



                          You want to reach a quadratic trinomial on the LHS so you can solve it. This is done by adding $big(frac{b}{2a}big)^2$ to both sides.



                          $$x^2+frac{b}{a}x+bigg(frac{b}{2a}bigg)^2 = -frac{c}{a}+frac{b^2}{4a^2}$$



                          $$bigg(x+frac{b}{2a}bigg)^2 = frac{b^2-4ac}{4a^2}$$



                          After which you can take the square root of both sides and solve for $x$. Can you use this general plan to solve the equation you’ve given?






                          share|cite|improve this answer
























                            2












                            2








                            2






                            In completing the square, you want to follow a plan like below.



                            $$ax^2+bx+c = 0$$



                            $$x^2+frac{b}{a}x+frac{c}{a} = 0 implies x^2+frac{b}{a}x = -frac{c}{a}$$



                            You want to reach a quadratic trinomial on the LHS so you can solve it. This is done by adding $big(frac{b}{2a}big)^2$ to both sides.



                            $$x^2+frac{b}{a}x+bigg(frac{b}{2a}bigg)^2 = -frac{c}{a}+frac{b^2}{4a^2}$$



                            $$bigg(x+frac{b}{2a}bigg)^2 = frac{b^2-4ac}{4a^2}$$



                            After which you can take the square root of both sides and solve for $x$. Can you use this general plan to solve the equation you’ve given?






                            share|cite|improve this answer












                            In completing the square, you want to follow a plan like below.



                            $$ax^2+bx+c = 0$$



                            $$x^2+frac{b}{a}x+frac{c}{a} = 0 implies x^2+frac{b}{a}x = -frac{c}{a}$$



                            You want to reach a quadratic trinomial on the LHS so you can solve it. This is done by adding $big(frac{b}{2a}big)^2$ to both sides.



                            $$x^2+frac{b}{a}x+bigg(frac{b}{2a}bigg)^2 = -frac{c}{a}+frac{b^2}{4a^2}$$



                            $$bigg(x+frac{b}{2a}bigg)^2 = frac{b^2-4ac}{4a^2}$$



                            After which you can take the square root of both sides and solve for $x$. Can you use this general plan to solve the equation you’ve given?







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 24 at 11:22









                            KM101

                            3,958417




                            3,958417






























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