the product of two numbers is six times their sum. the sum of their squares is 325. what are the two numbers?
I'm assuming that you want an exact answer, not a graphical approximation (which is a lost easier!):
x² + y² = 325 ===> x² + 2xy + y² = 325 + 2xy ===> (x + y)² = 325 + 2·6(x + y)
===> (x + y)² = 325 + 12(x + y) ===> (x + y)² - 12(x + y) - 325 = 0
Factor or use the quadratic formula: [(x + y) - 25][(x + y ) + 13] = 0
===> x = 25 or x = -13
If x + y = 25 ===> y = 25 - x
x² + y² = 325 ===> x² + (25 - x)² = 325 ===> x² + 625 - 50x + x² = 325
===> 2x² - 50x + 300 = 0 ===> x² - 25x + 150 = 0
Factoring: (x - 15)(x - 10) = 0 ===> x = 15 or x = 10
===> when x = 10, y = 15; when x = 15, y = 10
Now, x = -13 gives you two more answers; in a similar solution as the above: x = (-13 ± √481 ) / 2
When x = (-13 + √481 ) / 2, y = (-13 - √481 ) / 2
and when x = (-13 - √481 ) / 2, y = (-13 + √481 ) / 2
I'm assuming that you want an exact answer, not a graphical approximation (which is a lost easier!):
x² + y² = 325 ===> x² + 2xy + y² = 325 + 2xy ===> (x + y)² = 325 + 2·6(x + y)
===> (x + y)² = 325 + 12(x + y) ===> (x + y)² - 12(x + y) - 325 = 0
Factor or use the quadratic formula: [(x + y) - 25][(x + y ) + 13] = 0
===> x = 25 or x = -13
If x + y = 25 ===> y = 25 - x
x² + y² = 325 ===> x² + (25 - x)² = 325 ===> x² + 625 - 50x + x² = 325
===> 2x² - 50x + 300 = 0 ===> x² - 25x + 150 = 0
Factoring: (x - 15)(x - 10) = 0 ===> x = 15 or x = 10
===> when x = 10, y = 15; when x = 15, y = 10
Now, x = -13 gives you two more answers; in a similar solution as the above: x = (-13 ± √481 ) / 2
When x = (-13 + √481 ) / 2, y = (-13 - √481 ) / 2
and when x = (-13 - √481 ) / 2, y = (-13 + √481 ) / 2