Applying quadratic equations.












0












$begingroup$


So I realise this is quite an easy question, but for some reason I can't see the solution. So the question followed on from a previous question where we used a quadratic equation to find the dimensions of a right angle triangle.



This question is: Find the dimensions of all rectangles in which the area equals the perimeter + $3.5$, and in which the longer sides are twice the length of the shorter sides.



So my solution is:



side = $x$

length = $2x$



so then:
length x width = length + length + width + width + $3.5 $



$2x*x = 2x + 2x + x + x + 3.5 $



$2x^2 = 6x + 3.5 $



$-2x^2 + 6x + 3.5 = 0$



Then I would factorise this to find the positive values of $x$ that can be then used to determine the length of the rectangle. I am struggling to factorise while one of the values is $3.5$. Is there a way to more easily conceptualise how to factorise with non whole values? thanks.






quadratics







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Multiply by $2$
    $endgroup$
    – Claude Leibovici
    Jul 14 '18 at 6:14










  • $begingroup$
    Do you mean multiply every term by two,or? Because multiplying 3.5 gives 7, not six.
    $endgroup$
    – MEcho
    Jul 14 '18 at 6:16










  • $begingroup$
    Note that $2cdot 7-2cdot 1=12$... (Yes, you would multiply through the whole quadratic by $2$.)
    $endgroup$
    – abiessu
    Jul 14 '18 at 6:20












  • $begingroup$
    Oh okay, I see now! thanks!
    $endgroup$
    – MEcho
    Jul 14 '18 at 6:21










  • $begingroup$
    Use the quadratic formula.
    $endgroup$
    – MalayTheDynamo
    Jul 14 '18 at 7:06
















0












$begingroup$


So I realise this is quite an easy question, but for some reason I can't see the solution. So the question followed on from a previous question where we used a quadratic equation to find the dimensions of a right angle triangle.



This question is: Find the dimensions of all rectangles in which the area equals the perimeter + $3.5$, and in which the longer sides are twice the length of the shorter sides.



So my solution is:



side = $x$

length = $2x$



so then:
length x width = length + length + width + width + $3.5 $



$2x*x = 2x + 2x + x + x + 3.5 $



$2x^2 = 6x + 3.5 $



$-2x^2 + 6x + 3.5 = 0$



Then I would factorise this to find the positive values of $x$ that can be then used to determine the length of the rectangle. I am struggling to factorise while one of the values is $3.5$. Is there a way to more easily conceptualise how to factorise with non whole values? thanks.






quadratics







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Multiply by $2$
    $endgroup$
    – Claude Leibovici
    Jul 14 '18 at 6:14










  • $begingroup$
    Do you mean multiply every term by two,or? Because multiplying 3.5 gives 7, not six.
    $endgroup$
    – MEcho
    Jul 14 '18 at 6:16










  • $begingroup$
    Note that $2cdot 7-2cdot 1=12$... (Yes, you would multiply through the whole quadratic by $2$.)
    $endgroup$
    – abiessu
    Jul 14 '18 at 6:20












  • $begingroup$
    Oh okay, I see now! thanks!
    $endgroup$
    – MEcho
    Jul 14 '18 at 6:21










  • $begingroup$
    Use the quadratic formula.
    $endgroup$
    – MalayTheDynamo
    Jul 14 '18 at 7:06














0












0








0





$begingroup$


So I realise this is quite an easy question, but for some reason I can't see the solution. So the question followed on from a previous question where we used a quadratic equation to find the dimensions of a right angle triangle.



This question is: Find the dimensions of all rectangles in which the area equals the perimeter + $3.5$, and in which the longer sides are twice the length of the shorter sides.



So my solution is:



side = $x$

length = $2x$



so then:
length x width = length + length + width + width + $3.5 $



$2x*x = 2x + 2x + x + x + 3.5 $



$2x^2 = 6x + 3.5 $



$-2x^2 + 6x + 3.5 = 0$



Then I would factorise this to find the positive values of $x$ that can be then used to determine the length of the rectangle. I am struggling to factorise while one of the values is $3.5$. Is there a way to more easily conceptualise how to factorise with non whole values? thanks.






quadratics







share|cite|improve this question











$endgroup$




So I realise this is quite an easy question, but for some reason I can't see the solution. So the question followed on from a previous question where we used a quadratic equation to find the dimensions of a right angle triangle.



This question is: Find the dimensions of all rectangles in which the area equals the perimeter + $3.5$, and in which the longer sides are twice the length of the shorter sides.



So my solution is:



side = $x$

length = $2x$



so then:
length x width = length + length + width + width + $3.5 $



$2x*x = 2x + 2x + x + x + 3.5 $



$2x^2 = 6x + 3.5 $



$-2x^2 + 6x + 3.5 = 0$



Then I would factorise this to find the positive values of $x$ that can be then used to determine the length of the rectangle. I am struggling to factorise while one of the values is $3.5$. Is there a way to more easily conceptualise how to factorise with non whole values? thanks.






quadratics




quadratics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jul 14 '18 at 6:17







user570193

















asked Jul 14 '18 at 6:10









MEchoMEcho

11




11








  • 4




    $begingroup$
    Multiply by $2$
    $endgroup$
    – Claude Leibovici
    Jul 14 '18 at 6:14










  • $begingroup$
    Do you mean multiply every term by two,or? Because multiplying 3.5 gives 7, not six.
    $endgroup$
    – MEcho
    Jul 14 '18 at 6:16










  • $begingroup$
    Note that $2cdot 7-2cdot 1=12$... (Yes, you would multiply through the whole quadratic by $2$.)
    $endgroup$
    – abiessu
    Jul 14 '18 at 6:20












  • $begingroup$
    Oh okay, I see now! thanks!
    $endgroup$
    – MEcho
    Jul 14 '18 at 6:21










  • $begingroup$
    Use the quadratic formula.
    $endgroup$
    – MalayTheDynamo
    Jul 14 '18 at 7:06














  • 4




    $begingroup$
    Multiply by $2$
    $endgroup$
    – Claude Leibovici
    Jul 14 '18 at 6:14










  • $begingroup$
    Do you mean multiply every term by two,or? Because multiplying 3.5 gives 7, not six.
    $endgroup$
    – MEcho
    Jul 14 '18 at 6:16










  • $begingroup$
    Note that $2cdot 7-2cdot 1=12$... (Yes, you would multiply through the whole quadratic by $2$.)
    $endgroup$
    – abiessu
    Jul 14 '18 at 6:20












  • $begingroup$
    Oh okay, I see now! thanks!
    $endgroup$
    – MEcho
    Jul 14 '18 at 6:21










  • $begingroup$
    Use the quadratic formula.
    $endgroup$
    – MalayTheDynamo
    Jul 14 '18 at 7:06








4




4




$begingroup$
Multiply by $2$
$endgroup$
– Claude Leibovici
Jul 14 '18 at 6:14




$begingroup$
Multiply by $2$
$endgroup$
– Claude Leibovici
Jul 14 '18 at 6:14












$begingroup$
Do you mean multiply every term by two,or? Because multiplying 3.5 gives 7, not six.
$endgroup$
– MEcho
Jul 14 '18 at 6:16




$begingroup$
Do you mean multiply every term by two,or? Because multiplying 3.5 gives 7, not six.
$endgroup$
– MEcho
Jul 14 '18 at 6:16












$begingroup$
Note that $2cdot 7-2cdot 1=12$... (Yes, you would multiply through the whole quadratic by $2$.)
$endgroup$
– abiessu
Jul 14 '18 at 6:20






$begingroup$
Note that $2cdot 7-2cdot 1=12$... (Yes, you would multiply through the whole quadratic by $2$.)
$endgroup$
– abiessu
Jul 14 '18 at 6:20














$begingroup$
Oh okay, I see now! thanks!
$endgroup$
– MEcho
Jul 14 '18 at 6:21




$begingroup$
Oh okay, I see now! thanks!
$endgroup$
– MEcho
Jul 14 '18 at 6:21












$begingroup$
Use the quadratic formula.
$endgroup$
– MalayTheDynamo
Jul 14 '18 at 7:06




$begingroup$
Use the quadratic formula.
$endgroup$
– MalayTheDynamo
Jul 14 '18 at 7:06










3 Answers
3






active

oldest

votes


















0












$begingroup$

"Is there a way to more easily conceptualise how to factorise with non whole values? thanks. "



Yeah, it's called completing the square or the quadratic formula.



If you have $ax^2 + bx + c = 0$ and $a ne 0$ then.....



$ax^2 + bx + c = 0$



$x^2 + frac ba = -frac ca$



$x^2 + 2frac b{2a} = - frac ca$



$x^2 + 2frac b{2a} + (frac b{2a})^2 = (frac b{2a})^2 - frac ca$



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



$x + frac b{2a} = pm sqrt {frac {b^2 - 4ac}{4a^2}} = frac {sqrt{b^2 - 4ac}}{2a}$



$x = frac {-b pm sqrt {b^2 - 4ac}}{2a}$.



So in your case $x = frac {-6 pm sqrt{6^2 - 4*3.5*(-2)}}{2*(-2)} $



$= frac {-6 pm sqrt{36 + 28}}{-4}$



$= frac {6 mp sqrt{64}}{4} $



$= frac {6 mp 8}{4} $



$-frac 12$ or $frac 72$. As $x > 0$ we have $x = frac 72$ so



the sides are $frac 72$ and $7$ and area is $frac 72*7 = frac {49}2 = 24frac 12$ and the perimeter is $frac 72 + frac 72 + 7 + 7 = 21$.



And .... indeed, $24frac 12 = 21 + 3.5$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Without applying (or re-inventing) the formulæ, multiply your equation by $2 $ to rewrite it as
    $$4x^2-12x-7=(2x-3)^2-9-7=0iff (2x-3)^2=4^2$$
    so the longer size is $2x=4+3=7:$ and the shorter size is $:x=7/2$.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      $$2x^2 - 6x - 3.5 = 0$$



      I wondered if the $ac$ method would still work. It sort of can.



      $ac =2(-3.5) = -7$ and $b=-6$.



      We note that $(-7)(1) = -7 = ac$ and $(-7)+(1) = -6 = b$.



      So we replace $bx=-6x$ with $-6x = -7x + 1x$.



      begin{align}
      2x^2 - 6x - 3.5
      &= 2x^2 -7x + 1x - 3.5 \
      &= 2x(x-3.5) + 1(x-3.5) \
      &= (2x+1)(x-3.5) \
      hline
      x &in {-0.5, 3.5}
      end{align}






      share|cite|improve this answer











      $endgroup$













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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









        0












        $begingroup$

        "Is there a way to more easily conceptualise how to factorise with non whole values? thanks. "



        Yeah, it's called completing the square or the quadratic formula.



        If you have $ax^2 + bx + c = 0$ and $a ne 0$ then.....



        $ax^2 + bx + c = 0$



        $x^2 + frac ba = -frac ca$



        $x^2 + 2frac b{2a} = - frac ca$



        $x^2 + 2frac b{2a} + (frac b{2a})^2 = (frac b{2a})^2 - frac ca$



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



        $x + frac b{2a} = pm sqrt {frac {b^2 - 4ac}{4a^2}} = frac {sqrt{b^2 - 4ac}}{2a}$



        $x = frac {-b pm sqrt {b^2 - 4ac}}{2a}$.



        So in your case $x = frac {-6 pm sqrt{6^2 - 4*3.5*(-2)}}{2*(-2)} $



        $= frac {-6 pm sqrt{36 + 28}}{-4}$



        $= frac {6 mp sqrt{64}}{4} $



        $= frac {6 mp 8}{4} $



        $-frac 12$ or $frac 72$. As $x > 0$ we have $x = frac 72$ so



        the sides are $frac 72$ and $7$ and area is $frac 72*7 = frac {49}2 = 24frac 12$ and the perimeter is $frac 72 + frac 72 + 7 + 7 = 21$.



        And .... indeed, $24frac 12 = 21 + 3.5$.






        share|cite|improve this answer









        $endgroup$


















          0












          $begingroup$

          "Is there a way to more easily conceptualise how to factorise with non whole values? thanks. "



          Yeah, it's called completing the square or the quadratic formula.



          If you have $ax^2 + bx + c = 0$ and $a ne 0$ then.....



          $ax^2 + bx + c = 0$



          $x^2 + frac ba = -frac ca$



          $x^2 + 2frac b{2a} = - frac ca$



          $x^2 + 2frac b{2a} + (frac b{2a})^2 = (frac b{2a})^2 - frac ca$



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



          $x + frac b{2a} = pm sqrt {frac {b^2 - 4ac}{4a^2}} = frac {sqrt{b^2 - 4ac}}{2a}$



          $x = frac {-b pm sqrt {b^2 - 4ac}}{2a}$.



          So in your case $x = frac {-6 pm sqrt{6^2 - 4*3.5*(-2)}}{2*(-2)} $



          $= frac {-6 pm sqrt{36 + 28}}{-4}$



          $= frac {6 mp sqrt{64}}{4} $



          $= frac {6 mp 8}{4} $



          $-frac 12$ or $frac 72$. As $x > 0$ we have $x = frac 72$ so



          the sides are $frac 72$ and $7$ and area is $frac 72*7 = frac {49}2 = 24frac 12$ and the perimeter is $frac 72 + frac 72 + 7 + 7 = 21$.



          And .... indeed, $24frac 12 = 21 + 3.5$.






          share|cite|improve this answer









          $endgroup$
















            0












            0








            0





            $begingroup$

            "Is there a way to more easily conceptualise how to factorise with non whole values? thanks. "



            Yeah, it's called completing the square or the quadratic formula.



            If you have $ax^2 + bx + c = 0$ and $a ne 0$ then.....



            $ax^2 + bx + c = 0$



            $x^2 + frac ba = -frac ca$



            $x^2 + 2frac b{2a} = - frac ca$



            $x^2 + 2frac b{2a} + (frac b{2a})^2 = (frac b{2a})^2 - frac ca$



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



            $x + frac b{2a} = pm sqrt {frac {b^2 - 4ac}{4a^2}} = frac {sqrt{b^2 - 4ac}}{2a}$



            $x = frac {-b pm sqrt {b^2 - 4ac}}{2a}$.



            So in your case $x = frac {-6 pm sqrt{6^2 - 4*3.5*(-2)}}{2*(-2)} $



            $= frac {-6 pm sqrt{36 + 28}}{-4}$



            $= frac {6 mp sqrt{64}}{4} $



            $= frac {6 mp 8}{4} $



            $-frac 12$ or $frac 72$. As $x > 0$ we have $x = frac 72$ so



            the sides are $frac 72$ and $7$ and area is $frac 72*7 = frac {49}2 = 24frac 12$ and the perimeter is $frac 72 + frac 72 + 7 + 7 = 21$.



            And .... indeed, $24frac 12 = 21 + 3.5$.






            share|cite|improve this answer









            $endgroup$



            "Is there a way to more easily conceptualise how to factorise with non whole values? thanks. "



            Yeah, it's called completing the square or the quadratic formula.



            If you have $ax^2 + bx + c = 0$ and $a ne 0$ then.....



            $ax^2 + bx + c = 0$



            $x^2 + frac ba = -frac ca$



            $x^2 + 2frac b{2a} = - frac ca$



            $x^2 + 2frac b{2a} + (frac b{2a})^2 = (frac b{2a})^2 - frac ca$



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



            $x + frac b{2a} = pm sqrt {frac {b^2 - 4ac}{4a^2}} = frac {sqrt{b^2 - 4ac}}{2a}$



            $x = frac {-b pm sqrt {b^2 - 4ac}}{2a}$.



            So in your case $x = frac {-6 pm sqrt{6^2 - 4*3.5*(-2)}}{2*(-2)} $



            $= frac {-6 pm sqrt{36 + 28}}{-4}$



            $= frac {6 mp sqrt{64}}{4} $



            $= frac {6 mp 8}{4} $



            $-frac 12$ or $frac 72$. As $x > 0$ we have $x = frac 72$ so



            the sides are $frac 72$ and $7$ and area is $frac 72*7 = frac {49}2 = 24frac 12$ and the perimeter is $frac 72 + frac 72 + 7 + 7 = 21$.



            And .... indeed, $24frac 12 = 21 + 3.5$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jul 14 '18 at 7:27









            fleabloodfleablood

            72k22687




            72k22687























                0












                $begingroup$

                Without applying (or re-inventing) the formulæ, multiply your equation by $2 $ to rewrite it as
                $$4x^2-12x-7=(2x-3)^2-9-7=0iff (2x-3)^2=4^2$$
                so the longer size is $2x=4+3=7:$ and the shorter size is $:x=7/2$.






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  Without applying (or re-inventing) the formulæ, multiply your equation by $2 $ to rewrite it as
                  $$4x^2-12x-7=(2x-3)^2-9-7=0iff (2x-3)^2=4^2$$
                  so the longer size is $2x=4+3=7:$ and the shorter size is $:x=7/2$.






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    Without applying (or re-inventing) the formulæ, multiply your equation by $2 $ to rewrite it as
                    $$4x^2-12x-7=(2x-3)^2-9-7=0iff (2x-3)^2=4^2$$
                    so the longer size is $2x=4+3=7:$ and the shorter size is $:x=7/2$.






                    share|cite|improve this answer









                    $endgroup$



                    Without applying (or re-inventing) the formulæ, multiply your equation by $2 $ to rewrite it as
                    $$4x^2-12x-7=(2x-3)^2-9-7=0iff (2x-3)^2=4^2$$
                    so the longer size is $2x=4+3=7:$ and the shorter size is $:x=7/2$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jul 14 '18 at 8:01









                    BernardBernard

                    122k741116




                    122k741116























                        0












                        $begingroup$

                        $$2x^2 - 6x - 3.5 = 0$$



                        I wondered if the $ac$ method would still work. It sort of can.



                        $ac =2(-3.5) = -7$ and $b=-6$.



                        We note that $(-7)(1) = -7 = ac$ and $(-7)+(1) = -6 = b$.



                        So we replace $bx=-6x$ with $-6x = -7x + 1x$.



                        begin{align}
                        2x^2 - 6x - 3.5
                        &= 2x^2 -7x + 1x - 3.5 \
                        &= 2x(x-3.5) + 1(x-3.5) \
                        &= (2x+1)(x-3.5) \
                        hline
                        x &in {-0.5, 3.5}
                        end{align}






                        share|cite|improve this answer











                        $endgroup$


















                          0












                          $begingroup$

                          $$2x^2 - 6x - 3.5 = 0$$



                          I wondered if the $ac$ method would still work. It sort of can.



                          $ac =2(-3.5) = -7$ and $b=-6$.



                          We note that $(-7)(1) = -7 = ac$ and $(-7)+(1) = -6 = b$.



                          So we replace $bx=-6x$ with $-6x = -7x + 1x$.



                          begin{align}
                          2x^2 - 6x - 3.5
                          &= 2x^2 -7x + 1x - 3.5 \
                          &= 2x(x-3.5) + 1(x-3.5) \
                          &= (2x+1)(x-3.5) \
                          hline
                          x &in {-0.5, 3.5}
                          end{align}






                          share|cite|improve this answer











                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            $$2x^2 - 6x - 3.5 = 0$$



                            I wondered if the $ac$ method would still work. It sort of can.



                            $ac =2(-3.5) = -7$ and $b=-6$.



                            We note that $(-7)(1) = -7 = ac$ and $(-7)+(1) = -6 = b$.



                            So we replace $bx=-6x$ with $-6x = -7x + 1x$.



                            begin{align}
                            2x^2 - 6x - 3.5
                            &= 2x^2 -7x + 1x - 3.5 \
                            &= 2x(x-3.5) + 1(x-3.5) \
                            &= (2x+1)(x-3.5) \
                            hline
                            x &in {-0.5, 3.5}
                            end{align}






                            share|cite|improve this answer











                            $endgroup$



                            $$2x^2 - 6x - 3.5 = 0$$



                            I wondered if the $ac$ method would still work. It sort of can.



                            $ac =2(-3.5) = -7$ and $b=-6$.



                            We note that $(-7)(1) = -7 = ac$ and $(-7)+(1) = -6 = b$.



                            So we replace $bx=-6x$ with $-6x = -7x + 1x$.



                            begin{align}
                            2x^2 - 6x - 3.5
                            &= 2x^2 -7x + 1x - 3.5 \
                            &= 2x(x-3.5) + 1(x-3.5) \
                            &= (2x+1)(x-3.5) \
                            hline
                            x &in {-0.5, 3.5}
                            end{align}







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Jan 3 at 13:55

























                            answered Jan 3 at 7:37









                            steven gregorysteven gregory

                            18.3k32258




                            18.3k32258






























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