Set up the given triangle on x-y coordinates with right angle at (0,0). So the two vertices are at (5,0) and (0,2
![sqrt{x} n]{3}](https://tex.z-dn.net/?f=sqrt%7Bx%7D%20n%5D%7B3%7D%20)
)
let (a,0) and (0,b) be two vertices of the<span> equilateral triangle. So the third vertex must be at </span>

for a pt (x,y) on line sx+ty=1, the minimum of

equals to

smallest value happens at

so area is

hence m=75, n=67, p=3
m+n+p = 75+67+3 = 145
Answer:
z = x^3 +1
Step-by-step explanation:
Noting the squared term, it makes sense to substitute for that term:
z = x^3 +1
gives ...
16z^2 -22z -3 = 0 . . . . the quadratic you want
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<em>Solutions derived from that substitution</em>
Factoring gives ...
16z^2 -24z +2z -3 = 0
8z(2z -3) +1(2z -3) = 0
(8z +1)(2z -3) = 0
z = -1/8 or 3/2
Then we can find x:
x^3 +1 = -1/8
x^3 = -9/8 . . . . . subtract 1
x = (-1/2)∛9 . . . . . one of the real solutions
__
x^3 +1 = 3/2
x^3 = 1/2 = 4/8 . . . . . . subtract 1
x = (1/2)∛4 . . . . . . the other real solution
The complex solutions will be the two complex cube roots of -9/8 and the two complex cube roots of 1/2.
Step-by-step explanation:


35/8 = (4 * 8 = 32) and 3 left over so answer is 4 3/8