SOLUTION: pleae answe these qustion 1.find two numbers whose sum is 13 and whose product is 40 2.the area of a rectangle is 94sq. units and its perimeter is 38units. find the length an

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: pleae answe these qustion 1.find two numbers whose sum is 13 and whose product is 40 2.the area of a rectangle is 94sq. units and its perimeter is 38units. find the length an      Log On


   

Question 874959: pleae answe these qustion
1.find two numbers whose sum is 13 and whose product is 40
2.the area of a rectangle is 94sq. units and its perimeter is 38units.
find the length and width of the rectangle.
thanks
Found 2 solutions by Alan3354, KMST:
Answer by Alan3354(69443) About Me  (Show Source):
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
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%2A1 and 40%2B1=41%3C%3E13 so it is not 40+and+%7B%7B%7B1
40=20%2A2 and 20%2B2=22%3C%3E13 so it is not 20+and+%7B%7B%7B2
40=10%2A4 and 10%2B4=14%3C%3E13 so it is not 10+and+%7B%7B%7B4
40=8%2A5 and 8%2B5=13 so the answer is highlight%288%29 and highlight%2813%29 .
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%2By=13--->y=13-x
So the product of the two numbers is
40=x%2813-x%29
40=13x-x%5E2
highlight%28x%5E2-13x%2B40=0%29
I see 3 ways to solve that quadratic equation.
1) I would factor it into %28x-8%29%28x-5%29=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%29--->highlight%28x=5%29-->y=13-x=13-5--->highlight%28y=8%29
x-8=0%29--->highlight%28x=8%29-->y=13-x=13-8--->highlight%28y=5%29
Either solution tells us that the two numbers are highlight%288%29 and highlight%285%29 .
2) we could use the quadratic formula that says that the solutions to
ax%5E2%2Bbx%2Bc=0 are given by
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
For x%5E2-13x%2B40=0 , a=1 , b=-13 , and c=40 , so

That gives us the answers
x=%2813%2B3%29%2F2--->x=16%2F2--->highlight%28x=8%29 and
x=%2813-3%29%2F2--->x=10%2F2--->highlight%28x=5%29 as above.
3) The third option is completing the square:
x%5E2-13x%2B40=0
x%5E2-13x=-40
x%5E2-13x%2B%2813%2F2%29%5E2=-40%2B%2813%2F2%29%5E2
%28x-13%2F2%29%5E2=-40%2B169%2F4
%28x-13%2F2%29%5E2=-160%2F4%2B169%2F4
%28x-13%2F2%29%5E2=9%2F4
%28x-13%2F2%29%5E2=%283%2F2%29%5E2
So either x-13%2F2=3%2F2--->x=13%2F2%2B3%2F2--->x=16%2F2--->highlight%28x=8%29 or
x-13%2F2=-3%2F2--->x=13%2F2-3%2F2--->x=10%2F2--->highlight%28x=5%29 .
Any way you solve the quadratic equation, the two numbers are highlight%288%29 and highlight%285%29 .

2. You have to find two numbers that are the measurements of length and width for that rectangle.
area=length%2Awidth
so length%2Awidth=94
perimeter=2%28length%2Bwidth%29
so 2%28length%2Bwidth%29=38
length%2Bwidth=38%2F2
length%2Bwidth=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%2F4-->x=9.5 units
The area of that square would be x%2Ax=9.5%2A9.5=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%2819-x%29--> 94=19x-x%5E2-->highlight%28x%5E2-19x%2B94=0%29
To apply the quadratic formula,
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+ ,
for this equation a=1 , b=-19 , and c=94 .
Applying the quadratic formula,
x+=+%28-%28-19%29+%2B-+sqrt%28%28-19%29%5E2-4%2A1%2A94+%29%29%2F%282%2A1%29+
x+=+%2819+%2B-+sqrt%28361-376%29%29%2F2
x+=+%2819+%2B-+sqrt%28-15%29%29%2F2
We have no real solutions, because sqrt%28-15%29 is no real number,
so there are no real solutions, but we could ask that fifth grader about imaginary ones.