Age problem-PLEASE HELP [on hold]












0














Hannah. added one to her age, then multiplied the result by $4$. She squared that product, then increased the result by $3$, then doubled it. The final result was $1214$ more than $60$ times her age. Find Hannah.'s age, which we can assume is a positive number. I simplified to $32x^2+4x-1179=0$ but I don't really know what to do now. Please help!










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put on hold as off-topic by Henrik, Xander Henderson, Nosrati, Paul Frost, Cesareo Jan 4 at 12:52


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Henrik, Xander Henderson, Nosrati, Paul Frost, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.













  • It's a quadratic equation. Solve it.
    – harshit54
    Jan 4 at 6:28










  • Quadratic formula?
    – Thomas Shelby
    Jan 4 at 6:28






  • 2




    Also this is not a linear-algebra problem.
    – harshit54
    Jan 4 at 6:28






  • 1




    Assuming your work done so far was correct, you are left with an equation of the form $Ax^2+ Bx + C = 0$, to which the possible values for $x$ would be $dfrac{-B+sqrt{B^2-4AC}}{2A}$ and $dfrac{-B-sqrt{B^2-4AC}}{2A}$. Find those values by plugging in the appropriate values for $A,B,C$ and note that people cannot be a negative number of years old (and still be capable of arithmetic).
    – JMoravitz
    Jan 4 at 6:30








  • 1




    Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
    – Shaun
    Jan 4 at 6:45
















0














Hannah. added one to her age, then multiplied the result by $4$. She squared that product, then increased the result by $3$, then doubled it. The final result was $1214$ more than $60$ times her age. Find Hannah.'s age, which we can assume is a positive number. I simplified to $32x^2+4x-1179=0$ but I don't really know what to do now. Please help!










share|cite|improve this question















put on hold as off-topic by Henrik, Xander Henderson, Nosrati, Paul Frost, Cesareo Jan 4 at 12:52


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Henrik, Xander Henderson, Nosrati, Paul Frost, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.













  • It's a quadratic equation. Solve it.
    – harshit54
    Jan 4 at 6:28










  • Quadratic formula?
    – Thomas Shelby
    Jan 4 at 6:28






  • 2




    Also this is not a linear-algebra problem.
    – harshit54
    Jan 4 at 6:28






  • 1




    Assuming your work done so far was correct, you are left with an equation of the form $Ax^2+ Bx + C = 0$, to which the possible values for $x$ would be $dfrac{-B+sqrt{B^2-4AC}}{2A}$ and $dfrac{-B-sqrt{B^2-4AC}}{2A}$. Find those values by plugging in the appropriate values for $A,B,C$ and note that people cannot be a negative number of years old (and still be capable of arithmetic).
    – JMoravitz
    Jan 4 at 6:30








  • 1




    Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
    – Shaun
    Jan 4 at 6:45














0












0








0







Hannah. added one to her age, then multiplied the result by $4$. She squared that product, then increased the result by $3$, then doubled it. The final result was $1214$ more than $60$ times her age. Find Hannah.'s age, which we can assume is a positive number. I simplified to $32x^2+4x-1179=0$ but I don't really know what to do now. Please help!










share|cite|improve this question















Hannah. added one to her age, then multiplied the result by $4$. She squared that product, then increased the result by $3$, then doubled it. The final result was $1214$ more than $60$ times her age. Find Hannah.'s age, which we can assume is a positive number. I simplified to $32x^2+4x-1179=0$ but I don't really know what to do now. Please help!







quadratics






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edited Jan 4 at 6:42









Sauhard Sharma

773117




773117










asked Jan 4 at 6:25









J. DOEEJ. DOEE

16927




16927




put on hold as off-topic by Henrik, Xander Henderson, Nosrati, Paul Frost, Cesareo Jan 4 at 12:52


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Henrik, Xander Henderson, Nosrati, Paul Frost, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.




put on hold as off-topic by Henrik, Xander Henderson, Nosrati, Paul Frost, Cesareo Jan 4 at 12:52


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Henrik, Xander Henderson, Nosrati, Paul Frost, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.












  • It's a quadratic equation. Solve it.
    – harshit54
    Jan 4 at 6:28










  • Quadratic formula?
    – Thomas Shelby
    Jan 4 at 6:28






  • 2




    Also this is not a linear-algebra problem.
    – harshit54
    Jan 4 at 6:28






  • 1




    Assuming your work done so far was correct, you are left with an equation of the form $Ax^2+ Bx + C = 0$, to which the possible values for $x$ would be $dfrac{-B+sqrt{B^2-4AC}}{2A}$ and $dfrac{-B-sqrt{B^2-4AC}}{2A}$. Find those values by plugging in the appropriate values for $A,B,C$ and note that people cannot be a negative number of years old (and still be capable of arithmetic).
    – JMoravitz
    Jan 4 at 6:30








  • 1




    Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
    – Shaun
    Jan 4 at 6:45


















  • It's a quadratic equation. Solve it.
    – harshit54
    Jan 4 at 6:28










  • Quadratic formula?
    – Thomas Shelby
    Jan 4 at 6:28






  • 2




    Also this is not a linear-algebra problem.
    – harshit54
    Jan 4 at 6:28






  • 1




    Assuming your work done so far was correct, you are left with an equation of the form $Ax^2+ Bx + C = 0$, to which the possible values for $x$ would be $dfrac{-B+sqrt{B^2-4AC}}{2A}$ and $dfrac{-B-sqrt{B^2-4AC}}{2A}$. Find those values by plugging in the appropriate values for $A,B,C$ and note that people cannot be a negative number of years old (and still be capable of arithmetic).
    – JMoravitz
    Jan 4 at 6:30








  • 1




    Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
    – Shaun
    Jan 4 at 6:45
















It's a quadratic equation. Solve it.
– harshit54
Jan 4 at 6:28




It's a quadratic equation. Solve it.
– harshit54
Jan 4 at 6:28












Quadratic formula?
– Thomas Shelby
Jan 4 at 6:28




Quadratic formula?
– Thomas Shelby
Jan 4 at 6:28




2




2




Also this is not a linear-algebra problem.
– harshit54
Jan 4 at 6:28




Also this is not a linear-algebra problem.
– harshit54
Jan 4 at 6:28




1




1




Assuming your work done so far was correct, you are left with an equation of the form $Ax^2+ Bx + C = 0$, to which the possible values for $x$ would be $dfrac{-B+sqrt{B^2-4AC}}{2A}$ and $dfrac{-B-sqrt{B^2-4AC}}{2A}$. Find those values by plugging in the appropriate values for $A,B,C$ and note that people cannot be a negative number of years old (and still be capable of arithmetic).
– JMoravitz
Jan 4 at 6:30






Assuming your work done so far was correct, you are left with an equation of the form $Ax^2+ Bx + C = 0$, to which the possible values for $x$ would be $dfrac{-B+sqrt{B^2-4AC}}{2A}$ and $dfrac{-B-sqrt{B^2-4AC}}{2A}$. Find those values by plugging in the appropriate values for $A,B,C$ and note that people cannot be a negative number of years old (and still be capable of arithmetic).
– JMoravitz
Jan 4 at 6:30






1




1




Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
– Shaun
Jan 4 at 6:45




Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
– Shaun
Jan 4 at 6:45










4 Answers
4






active

oldest

votes


















2














Letting Hannah's age be $x$ we obtain that



$$2left( left(4(x+1)right)^2 +3 right)-1214=60x$$



$$32x^2+4x-1176=0$$



Using quadratic formula, we obtain $x=6$ (taking the positive result only).






share|cite|improve this answer





























    1














    Assuming that your result is correct, we are left with the quadratic equation $32x^2+4x-1179=0$. To solve for $x$, there are various techniques. The most common (and general) one would be completing the square, which will lead you to the general quadratic formula, allowing you to solve any quadratic equation of the form $ax^2+bx+c=0$ by
    $$x=frac1{2a}left(-bpmsqrt{b^2-4ac}right).$$






    share|cite|improve this answer





























      1














      Your equation is slightly incorrect. The actual equation is as follows(based on your question)



      $$32x^2 +4x -1176=0$$



      This equation can be factored using the quadratic formula



      $$x=frac{-4pm sqrt{16+4cdot1176cdot32}}{2cdot32}$$



      $$x=frac{-4pm sqrt{150544}}{64}$$



      $$x=frac{-4pm 388}{64}$$



      Since age can't be negative



      $$x=(-4+388)/64 = 6$$






      share|cite|improve this answer





























        0














        $32x^2 +4x -1176=0$ can be divided by $4$, resulting in $8x^2 + x - 294 = 0$.



        The $ac$ method requires us to find two integers whose product is
        $8 cdot (-294) = -2cdot2cdot 2 cdot 2 cdot 3 cdot 7 cdot 7$ and whose sum is $1$.



        We note that $2$ cannot be a factor of both integers since then their sum will be a multiple of $2$ and $1$ is not a multiple of $2$. Likewise, 7 cannot be a factor of both integers. This leads us to guess that the two integers are



        $$text{$-2cdot 2 cdot 2 cdot 2 cdot 3 = -48 qquad $ and $qquad 7 cdot 7 = 49$}$$



        Next we replace $x =1x$ with $1x = -48x + 49x$ and factor.



        begin{align}
        32x^2 +4x -1176
        &= 4(8x^2 + color{red}1x - 294) \
        &= 4(8x^2 - color{red}{48}x + color{red}{49}x - 294) \
        &= 4((8x^2 - 48x) + (49x - 294)) \
        &= 4((8x(x - 6) + 49(x - 6)) \
        &= 4(8x+49)(x - 6) \
        end{align}






        share|cite|improve this answer




























          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2














          Letting Hannah's age be $x$ we obtain that



          $$2left( left(4(x+1)right)^2 +3 right)-1214=60x$$



          $$32x^2+4x-1176=0$$



          Using quadratic formula, we obtain $x=6$ (taking the positive result only).






          share|cite|improve this answer


























            2














            Letting Hannah's age be $x$ we obtain that



            $$2left( left(4(x+1)right)^2 +3 right)-1214=60x$$



            $$32x^2+4x-1176=0$$



            Using quadratic formula, we obtain $x=6$ (taking the positive result only).






            share|cite|improve this answer
























              2












              2








              2






              Letting Hannah's age be $x$ we obtain that



              $$2left( left(4(x+1)right)^2 +3 right)-1214=60x$$



              $$32x^2+4x-1176=0$$



              Using quadratic formula, we obtain $x=6$ (taking the positive result only).






              share|cite|improve this answer












              Letting Hannah's age be $x$ we obtain that



              $$2left( left(4(x+1)right)^2 +3 right)-1214=60x$$



              $$32x^2+4x-1176=0$$



              Using quadratic formula, we obtain $x=6$ (taking the positive result only).







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 4 at 6:35









              Hugh EntwistleHugh Entwistle

              836217




              836217























                  1














                  Assuming that your result is correct, we are left with the quadratic equation $32x^2+4x-1179=0$. To solve for $x$, there are various techniques. The most common (and general) one would be completing the square, which will lead you to the general quadratic formula, allowing you to solve any quadratic equation of the form $ax^2+bx+c=0$ by
                  $$x=frac1{2a}left(-bpmsqrt{b^2-4ac}right).$$






                  share|cite|improve this answer


























                    1














                    Assuming that your result is correct, we are left with the quadratic equation $32x^2+4x-1179=0$. To solve for $x$, there are various techniques. The most common (and general) one would be completing the square, which will lead you to the general quadratic formula, allowing you to solve any quadratic equation of the form $ax^2+bx+c=0$ by
                    $$x=frac1{2a}left(-bpmsqrt{b^2-4ac}right).$$






                    share|cite|improve this answer
























                      1












                      1








                      1






                      Assuming that your result is correct, we are left with the quadratic equation $32x^2+4x-1179=0$. To solve for $x$, there are various techniques. The most common (and general) one would be completing the square, which will lead you to the general quadratic formula, allowing you to solve any quadratic equation of the form $ax^2+bx+c=0$ by
                      $$x=frac1{2a}left(-bpmsqrt{b^2-4ac}right).$$






                      share|cite|improve this answer












                      Assuming that your result is correct, we are left with the quadratic equation $32x^2+4x-1179=0$. To solve for $x$, there are various techniques. The most common (and general) one would be completing the square, which will lead you to the general quadratic formula, allowing you to solve any quadratic equation of the form $ax^2+bx+c=0$ by
                      $$x=frac1{2a}left(-bpmsqrt{b^2-4ac}right).$$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Jan 4 at 6:34









                      YiFanYiFan

                      2,6791422




                      2,6791422























                          1














                          Your equation is slightly incorrect. The actual equation is as follows(based on your question)



                          $$32x^2 +4x -1176=0$$



                          This equation can be factored using the quadratic formula



                          $$x=frac{-4pm sqrt{16+4cdot1176cdot32}}{2cdot32}$$



                          $$x=frac{-4pm sqrt{150544}}{64}$$



                          $$x=frac{-4pm 388}{64}$$



                          Since age can't be negative



                          $$x=(-4+388)/64 = 6$$






                          share|cite|improve this answer


























                            1














                            Your equation is slightly incorrect. The actual equation is as follows(based on your question)



                            $$32x^2 +4x -1176=0$$



                            This equation can be factored using the quadratic formula



                            $$x=frac{-4pm sqrt{16+4cdot1176cdot32}}{2cdot32}$$



                            $$x=frac{-4pm sqrt{150544}}{64}$$



                            $$x=frac{-4pm 388}{64}$$



                            Since age can't be negative



                            $$x=(-4+388)/64 = 6$$






                            share|cite|improve this answer
























                              1












                              1








                              1






                              Your equation is slightly incorrect. The actual equation is as follows(based on your question)



                              $$32x^2 +4x -1176=0$$



                              This equation can be factored using the quadratic formula



                              $$x=frac{-4pm sqrt{16+4cdot1176cdot32}}{2cdot32}$$



                              $$x=frac{-4pm sqrt{150544}}{64}$$



                              $$x=frac{-4pm 388}{64}$$



                              Since age can't be negative



                              $$x=(-4+388)/64 = 6$$






                              share|cite|improve this answer












                              Your equation is slightly incorrect. The actual equation is as follows(based on your question)



                              $$32x^2 +4x -1176=0$$



                              This equation can be factored using the quadratic formula



                              $$x=frac{-4pm sqrt{16+4cdot1176cdot32}}{2cdot32}$$



                              $$x=frac{-4pm sqrt{150544}}{64}$$



                              $$x=frac{-4pm 388}{64}$$



                              Since age can't be negative



                              $$x=(-4+388)/64 = 6$$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Jan 4 at 6:34









                              Sauhard SharmaSauhard Sharma

                              773117




                              773117























                                  0














                                  $32x^2 +4x -1176=0$ can be divided by $4$, resulting in $8x^2 + x - 294 = 0$.



                                  The $ac$ method requires us to find two integers whose product is
                                  $8 cdot (-294) = -2cdot2cdot 2 cdot 2 cdot 3 cdot 7 cdot 7$ and whose sum is $1$.



                                  We note that $2$ cannot be a factor of both integers since then their sum will be a multiple of $2$ and $1$ is not a multiple of $2$. Likewise, 7 cannot be a factor of both integers. This leads us to guess that the two integers are



                                  $$text{$-2cdot 2 cdot 2 cdot 2 cdot 3 = -48 qquad $ and $qquad 7 cdot 7 = 49$}$$



                                  Next we replace $x =1x$ with $1x = -48x + 49x$ and factor.



                                  begin{align}
                                  32x^2 +4x -1176
                                  &= 4(8x^2 + color{red}1x - 294) \
                                  &= 4(8x^2 - color{red}{48}x + color{red}{49}x - 294) \
                                  &= 4((8x^2 - 48x) + (49x - 294)) \
                                  &= 4((8x(x - 6) + 49(x - 6)) \
                                  &= 4(8x+49)(x - 6) \
                                  end{align}






                                  share|cite|improve this answer


























                                    0














                                    $32x^2 +4x -1176=0$ can be divided by $4$, resulting in $8x^2 + x - 294 = 0$.



                                    The $ac$ method requires us to find two integers whose product is
                                    $8 cdot (-294) = -2cdot2cdot 2 cdot 2 cdot 3 cdot 7 cdot 7$ and whose sum is $1$.



                                    We note that $2$ cannot be a factor of both integers since then their sum will be a multiple of $2$ and $1$ is not a multiple of $2$. Likewise, 7 cannot be a factor of both integers. This leads us to guess that the two integers are



                                    $$text{$-2cdot 2 cdot 2 cdot 2 cdot 3 = -48 qquad $ and $qquad 7 cdot 7 = 49$}$$



                                    Next we replace $x =1x$ with $1x = -48x + 49x$ and factor.



                                    begin{align}
                                    32x^2 +4x -1176
                                    &= 4(8x^2 + color{red}1x - 294) \
                                    &= 4(8x^2 - color{red}{48}x + color{red}{49}x - 294) \
                                    &= 4((8x^2 - 48x) + (49x - 294)) \
                                    &= 4((8x(x - 6) + 49(x - 6)) \
                                    &= 4(8x+49)(x - 6) \
                                    end{align}






                                    share|cite|improve this answer
























                                      0












                                      0








                                      0






                                      $32x^2 +4x -1176=0$ can be divided by $4$, resulting in $8x^2 + x - 294 = 0$.



                                      The $ac$ method requires us to find two integers whose product is
                                      $8 cdot (-294) = -2cdot2cdot 2 cdot 2 cdot 3 cdot 7 cdot 7$ and whose sum is $1$.



                                      We note that $2$ cannot be a factor of both integers since then their sum will be a multiple of $2$ and $1$ is not a multiple of $2$. Likewise, 7 cannot be a factor of both integers. This leads us to guess that the two integers are



                                      $$text{$-2cdot 2 cdot 2 cdot 2 cdot 3 = -48 qquad $ and $qquad 7 cdot 7 = 49$}$$



                                      Next we replace $x =1x$ with $1x = -48x + 49x$ and factor.



                                      begin{align}
                                      32x^2 +4x -1176
                                      &= 4(8x^2 + color{red}1x - 294) \
                                      &= 4(8x^2 - color{red}{48}x + color{red}{49}x - 294) \
                                      &= 4((8x^2 - 48x) + (49x - 294)) \
                                      &= 4((8x(x - 6) + 49(x - 6)) \
                                      &= 4(8x+49)(x - 6) \
                                      end{align}






                                      share|cite|improve this answer












                                      $32x^2 +4x -1176=0$ can be divided by $4$, resulting in $8x^2 + x - 294 = 0$.



                                      The $ac$ method requires us to find two integers whose product is
                                      $8 cdot (-294) = -2cdot2cdot 2 cdot 2 cdot 3 cdot 7 cdot 7$ and whose sum is $1$.



                                      We note that $2$ cannot be a factor of both integers since then their sum will be a multiple of $2$ and $1$ is not a multiple of $2$. Likewise, 7 cannot be a factor of both integers. This leads us to guess that the two integers are



                                      $$text{$-2cdot 2 cdot 2 cdot 2 cdot 3 = -48 qquad $ and $qquad 7 cdot 7 = 49$}$$



                                      Next we replace $x =1x$ with $1x = -48x + 49x$ and factor.



                                      begin{align}
                                      32x^2 +4x -1176
                                      &= 4(8x^2 + color{red}1x - 294) \
                                      &= 4(8x^2 - color{red}{48}x + color{red}{49}x - 294) \
                                      &= 4((8x^2 - 48x) + (49x - 294)) \
                                      &= 4((8x(x - 6) + 49(x - 6)) \
                                      &= 4(8x+49)(x - 6) \
                                      end{align}







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Jan 4 at 8:12









                                      steven gregorysteven gregory

                                      17.7k32257




                                      17.7k32257















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