how would i find the perimeter of a triangle with polynomials

Question

find a perimeter of a triangle with these sides x^2-6x+4 and x^2-7x+13.

Answer 1

The sides of triangle are x^2 -6x + 4 and x^2 -7x +13

We can't determine the perimeter of triangle with two sides.

Perimeter of triangle = a + b + c

If it is a right angle triangle

From the pythagorean therom

a^2 + b^2 = c^2

Find the length of the third side, then add the sides to find the perimeter.

Assume that, the triangle is a right triangle.

And the sides are a = x^2- 6x + 4 , b = x^2 - 7x + 13

(x^2 - 6x + 4)^2 + (x^2 - 7x + 13)^2 = c^2

x^4 - 12x^3 + 44x^2 - 48x + 16 + x^4 -14x^3 + 75x^2 - 182x + 169 = c^2

c^2 = 2x^4 - 26x^3  + 119x^2 - 230x + 185

c = sqrt(2x^4 - 26x^3  + 119x^2 - 230x + 185)

P = a + b + c

P = (x^2 - 6x + 4) + (x^2 - 7x + 13) + sqrt(2x^4 - 26x^3  + 119x^2 - 230x + 185)

P = 2x^2 -13x + 17 + sqrt(2x^4 - 26x^3  + 119x^2 - 230x + 185)