Question 874959
1. You can set up a quadratic equation and solve it, and that may be what is expected, but it is not the most efficient way.
THE FIFTH GRADER WAY:
The most effective way would be to find pairs of factors whose product is {{{40}}}
and see if the sum of one of those pairs is {{{13}}} :
{{{40=40*1}}} and {{{40+1=41<>13}}} so it is not {{{40 and {{{1}}}
{{{40=20*2}}} and {{{20+2=22<>13}}} so it is not {{{20 and {{{2}}}
{{{40=10*4}}} and {{{10+4=14<>13}}} so it is not {{{10 and {{{4}}}
{{{40=8*5}}} and {{{8+5=13}}} so the answer is {{{highlight(8)}}} and {{{highlight(13)}}} .
I did not even need to mention "quadratic equations" or know what the term means.
INVOKING QUADRATIC EQUATIONS:
{{{x}}}= one of the numbers
{{{y}}}= the other number
Since we know they add to {{{13}}} ,
{{{x+y=13}}}--->{{{y=13-x}}}
So the product of the two numbers is
{{{40=x(13-x)}}}
{{{40=13x-x^2}}}
{{{highlight(x^2-13x+40=0)}}}
I see 3 ways to solve that quadratic equation.
1) I would factor it into {{{(x-8)(x-5)=0}}}
and then make each factor equal to zero.
Factoring really means doing the work the fifth grader would do,
and I believe that is the most efficient way to solve this particular quadratic equation.
{{{x-5=0)}}}--->{{{highlight(x=5)}}}-->{{{y=13-x=13-5}}}--->{{{highlight(y=8)}}}
{{{x-8=0)}}}--->{{{highlight(x=8)}}}-->{{{y=13-x=13-8}}}--->{{{highlight(y=5)}}}
Either solution tells us that the two numbers are {{{highlight(8)}}} and {{{highlight(5)}}} .
2) we could use the quadratic formula that says that the solutions to
{{{ax^2+bx+c=0}}} are given by
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
For {{{x^2-13x+40=0}}} , {{{a=1}}} , {{{b=-13}}} , and {{{c=40}}} , so
{{{x=(-(-13) +- sqrt((-13)^2-4*1*40))/(2*1)=(13 +- sqrt(169-160))/2=(13 +- sqrt(9))/2=(13 +- 3)/2}}}
That gives us the answers
{{{x=(13+3)/2}}}--->{{{x=16/2}}}--->{{{highlight(x=8)}}} and
{{{x=(13-3)/2}}}--->{{{x=10/2}}}--->{{{highlight(x=5)}}} as above.
3) The third option is completing the square:
{{{x^2-13x+40=0}}}
{{{x^2-13x=-40}}}
{{{x^2-13x+(13/2)^2=-40+(13/2)^2}}}
{{{(x-13/2)^2=-40+169/4}}}
{{{(x-13/2)^2=-160/4+169/4}}}
{{{(x-13/2)^2=9/4}}}
{{{(x-13/2)^2=(3/2)^2}}}
So either {{{x-13/2=3/2}}}--->{{{x=13/2+3/2}}}--->{{{x=16/2}}}--->{{{highlight(x=8)}}} or
{{{x-13/2=-3/2}}}--->{{{x=13/2-3/2}}}--->{{{x=10/2}}}--->{{{highlight(x=5)}}} .
Any way you solve the quadratic equation, the two numbers are {{{highlight(8)}}} and {{{highlight(5)}}} .
 
2. You have to find two numbers that are the measurements of {{{length}}} and {{{width}}} for that rectangle.
{{{area=length*width}}}
so {{{length*width=94}}}
{{{perimeter=2(length+width)}}}
so {{{2(length+width)=38}}}
{{{length+width=38/2}}}
{{{length+width=19}}}
THE FIFTH GRADER WAY/WITHOUT MENTIONING QUADRATIC EQUATIONS:
Could it be a square?
If it was a square with a side length of {{{x}}} units,
The perimeter would be {{{4x=38}}}--->{{{x=38/4}}}-->{{{x=9.5}}} units
The area of that square would be {{{x*x=9.5*9.5=90.25}}} square units .
A square with a perimeter of {{{38}}} units has an area of {{{90.25}}} square units, and we are asked to find a larger rectangle with the same perimeter and an area of {{{94}}} square units .
There is something wrong with this problem.
In the real world there is no such a rectangle.
Maybe there was a typo, or maybe it is a trick question,
because we know, from all those fencing-a-square-pen problems, that the largest rectangle we can make with a given perimeter is a square.
There are no real length and width measurements for that rectangle.
Could it be that the measurements are imaginary numbers?
The fifth grader goes to the computer to do a search on "imaginary numbers."
 
THE QUADRATIC WAY:
So the sum of those two measures is {{{19}}} and the product is {{{94}}} .
This problem is similar to the one above, but factoring does not always work.
Sometimes factoring is complicated,
Sometimes the solutions are irrational numbers, so you would have to write a quadratic equation and try to solve it by using the quadratic formula or completing the square.
Sometimes there are no real solutions.
In all those cases, the quadratic formula (or completing the square) would tell us the answer.
If one of those measures is {{{x}}} , the other will be {{{19-x}}}
and {{{94=x(19-x)}}}--> {{{94=19x-x^2}}}-->{{{highlight(x^2-19x+94=0)}}}
To apply the quadratic formula,
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} ,
for this equation {{{a=1}}} , {{{b=-19}}} , and {{{c=94}}} .
Applying the quadratic formula,
{{{x = (-(-19) +- sqrt((-19)^2-4*1*94 ))/(2*1) }}}
{{{x = (19 +- sqrt(361-376))/2}}}
{{{x = (19 +- sqrt(-15))/2}}}
We have no real solutions, because {{{sqrt(-15)}}} is no real number,
so there are no real solutions, but we could ask that fifth grader about imaginary ones.