SOLUTION: If there is anyone that can help me solve this or I should say these problems it would be a huge help I have a exam that I'm studying for and cant spend any more time tring to get

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Question 201016: If there is anyone that can help me solve this or I should say these problems it would be a huge help I have a exam that I'm studying for and cant spend any more time tring to get the solutions. Thank you to anyone nice enough to help me out!
#1a)The sum of the squares of two numbers is 128. The product of the numbers is 64. Find the numbers? (answer with a exact solution and in ordered pairs)
#1b) The difference between the squares of two numbers is 3. Twice the square of the first number increased by the square of the second number is 9. Find the numbers? (please put your solutions in the form of an ordered pair(s)
#1c) A 150 ffet of fence is available to enclose a 1125 square foot region in the shape of adjoining squares. With the sides of length X and Y. The big square has sides of length X and the small square has sides of length Y. Find X and Y?
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
I'll do the first two to get you going in the right direction.


a)

"The sum of the squares of two numbers is 128" translates to x%5E2%2By%5E2=128


"The product of the numbers is 64" translates to xy=64


xy=64 Start with the second equation.


y=64%2Fx Divide both sides by "x".


x%5E2%2By%5E2=128 Move onto the first equation


x%5E2%2B%2864%2Fx%29%5E2=128 Plug in y=64%2Fx


x%5E2%2B4096%2Fx%5E2=128 Square 64%2Fx to get 4096%2Fx%5E2


x%5E4%2B4096=128x%5E2 Multiply EVERY term by the LCD x%5E2 to clear out the fractions.


x%5E4%2B4096-128x%5E2=0 Subtract 128x%5E2 from both sides.


x%5E4-128x%5E2%2B4096=0 Rearrange the terms.


Let z=x%5E2. So z%5E2=x%5E4


z%5E2-128z%2B4096=0 Replace x%5E4 with z%5E2. Replace x%5E2 with z


Notice that the quadratic z%5E2-128z%2B4096 is in the form of Az%5E2%2BBz%2BC where A=1, B=-128, and C=4096


Let's use the quadratic formula to solve for "z":


z+=+%28-B+%2B-+sqrt%28+B%5E2-4AC+%29%29%2F%282A%29 Start with the quadratic formula


z+=+%28-%28-128%29+%2B-+sqrt%28+%28-128%29%5E2-4%281%29%284096%29+%29%29%2F%282%281%29%29 Plug in A=1, B=-128, and C=4096


z+=+%28128+%2B-+sqrt%28+%28-128%29%5E2-4%281%29%284096%29+%29%29%2F%282%281%29%29 Negate -128 to get 128.


z+=+%28128+%2B-+sqrt%28+16384-4%281%29%284096%29+%29%29%2F%282%281%29%29 Square -128 to get 16384.


z+=+%28128+%2B-+sqrt%28+16384-16384+%29%29%2F%282%281%29%29 Multiply 4%281%29%284096%29 to get 16384


z+=+%28128+%2B-+sqrt%28+0+%29%29%2F%282%281%29%29 Subtract 16384 from 16384 to get 0


z+=+%28128+%2B-+sqrt%28+0+%29%29%2F%282%29 Multiply 2 and 1 to get 2.


z+=+%28128+%2B-+0%29%2F%282%29 Take the square root of 0 to get 0.


z+=+%28128+%2B+0%29%2F%282%29 or z+=+%28128+-+0%29%2F%282%29 Break up the expression.


z+=+%28128%29%2F%282%29 or z+=++%28128%29%2F%282%29 Combine like terms.


z+=+64 or z+=+64 Simplify.


So the only solution (in terms of "z") is z+=+64


Now recall that we let z=x%5E2. So this means that 64=x%5E2 and that x=8 or x=-8

--------------------------

Now let's find "y" when x=8:


y=64%2Fx Go back to the first isolated equation.


y=64%2F8 Plug in x=8


y=8 Reduce.


So when x=8, y=8 giving the ordered pair (8,8)



--------------------------

Now let's find "y" when x=-8:


y=64%2Fx Go back to the first isolated equation.


y=64%2F%28-8%29 Plug in x=-8


y=-8 Reduce.


So when x=-8, y=-8 giving the ordered pair (-8,-8)


===============================================================

Answer:

So the two ordered pair solutions are (8,8) and (-8,-8)




b)


"The difference between the squares of two numbers is 3" translates to x%5E2-y%5E2=3


"Twice the square of the first number increased by the square of the second number is 9" translates to 2x%5E2%2By%5E2=9


x%5E2-y%5E2=3 Start with the first equation.


x%5E2=3%2By%5E2 Add y%5E2 to both sides.


2x%5E2%2By%5E2=9 Move onto the second equation.


2%283%2By%5E2%29%2By%5E2=9 Plug in x%5E2=3%2By%5E2


6%2B2y%5E2%2By%5E2=9 Distribute


2y%5E2%2By%5E2=9-6 Subtract 6 from both sides.


3y%5E2=3 Combine like terms.


y%5E2=3%2F3 Divide both sides by 3.


y%5E2=1 Reduce


y=sqrt%281%29 or y=-sqrt%281%29 Take the square root of both sides (don't forget the "plus/minus")


y=1 or y=-1 Take the square root of 1 to get 1

--------------------------


Now that we know "y", we can use these values to find "x".


Let's find "x" when y=1:


x%5E2=3%2By%5E2 Start with the given equation.


x%5E2=3%2B%281%29%5E2 Plug in y=1


x%5E2=3%2B1 Square 1 to get 1


x%5E2=4 Combine like terms.


x=2 or x=-2 Take the square root of both sides


So the first two ordered pair solutions are (2,1) and (-2,1)


--------------------------

x%5E2=3%2By%5E2 Start with the given equation.


x%5E2=3%2B%28-1%29%5E2 Plug in y=-1


x%5E2=3%2B1 Square 1 to get 1


x%5E2=4 Combine like terms.


x=2 or x=-2 Take the square root of both sides


So the next two ordered pair solutions are (2,-1) and (-2,-1)

========================================================

Answer:


So the four ordered pair solutions are:

(2,1), (-2,1), (2,-1) and (-2,-1)




c)

I'll let you do this one on your own. Let me know if you have questions.


If you have any questions, email me at jim_thompson5910@hotmail.com.
Check out my website if you are interested in tutoring.