SOLUTION: The sum of two squares of two numbers is 128, the product of the two numbers is 64. what are the 2 numbers?

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Question 201238: The sum of two squares of two numbers is 128, the product of the two numbers is 64. what are the 2 numbers?
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
"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)



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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)


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Answer:

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



This means that the two numbers are:

8 and 8,

OR

-8 and -8