Answer:
You did not post the options, but i will try to answer this in a general way.
Because we have two solutions, i know that we are talking about quadratic equations, of the form of:
0 = a*x^2 + b*x + c.
There are two easy ways to see if the solutions of this equation are real or not.
1) look at the graph, if the graph touches the x-axis, then we have real solutions (if the graph does not touch the x-axis, we have complex solutions).
2) look at the determinant.
The determinant of a quadratic equation is:
D = b^2 - 4*a*c.
if D > 0, we have two real solutions.
if D = 0, we have one real solution (or two real solutions that are equal)
if D < 0, we have two complex solutions.
A right triangle's longest side is the hypotenuse
let x=longest, y=middle, and z=shortest
x=y+2
y=2z-1
therefore x=(2z-1)+2=2z+1
find z
z^2+y^2=x^2 by Pythagorean theorem
plug in x and y in terms of z
z^2+(2z-1)^2=(2z+1)^2
z^2+4z^2-4z+1=4z^2+4z+1
subtract the right-hand side's value from the left-hand side's
z^2-8z=0
z(z-8)=0
z=0, 8
z cannot be zero as the sides must have some value to it.
Therefore the shortest side is equal to 8
1, ___, ____, _____, _____, _____, ____, 29
29-1 = 28
28/7 = 4
1, 5, 9, 13, 17, 21, 25, 29
Just pick a number greater than 1 and that would be for P
