Question 77881
1. Find the length of a shorter leg of a right triangle 1. Find the length of a shorter leg of a right triangle if the longer leg is 10 miles more than the shorter leg and the hypotenuse is 10 miles less than twice the shorter leg.
:
Using pythag: a^2 + b^2 = c^2
:
Let x = a: the shorter leg
:
"longer leg is 10 miles more than the shorter leg"
b = (x+10)
:
"the hypotenuse is 10 miles less than twice the shorter leg."
c = (2x - 10)
:
a^2 + b^2 = c^2 is:
x^2 + (x+10)^2 = (2x-10)^2
:
x^2 + (x^2 + 20x + 100) = 4x^2 - 40x + 100
:
2x^2 + 20x + 100 = 4x^2 - 40x + 100
:
0 = 4x^2 - 2x^2 - 40x - 20x + 100 - 100
:
2x^2 - 60x = 0; a simple equation
:
Factors to:
2x(x - 30) = 0
:
2x = 0
and
x = 30 the length of the shorter side
:
Check; a^2 + b^2 = c^2:
30^2 + 40^2 = (60-10)^2
900 + 1600 = 50^2 = 2500
:
:
2. "The sum of two numbers is twenty five"
x + y = 25
y = (25 - x)
: 
"the sum of their squares is three hundred and five"
x^2 + y^2 = 305
: 
Find the number.
Substitute (25-x) for y in the 2nd equation:
x^2 + (25-x)^2 = 305
:
x^2 + 625 - 50x + x^2 = 305
:
2x^2 - 50x + 625 - 305 = 0
:
2x^2 - 50x + 320 = 0
;
There is no solution: Look at the discriminant of this equation:
 -b^2 - 4 * a * c
-50^2 - 4 * 2 * 320
2500 - 2560 = -60; <0 means x has no real solutions
:
If we graph the two equations:
y = 25-x
y = Sqrt(305 - x^2)
{{{ graph( 300, 200, -2, 20, -2, 20, 25-x, sqrt(305-x^2)) }}}
:
No point of intersection