Solve the expression [(a^2+b^2+c^2)*(a^2+b^2-c^2)]/(a^2+b^2-2ab), if the quadratic is x^2-5x+6=0,keeping in mind the general form of quadratic...

To compute the expression, we need the values of
coefficients of the quadratic a,b,c.

The general form of
the quadratic equation is ax^2 +bx + c =
0.

Comparing the given
equation with the general form, we'll identify a,b,c: a = 1, b = -5 and c =
6

We recognize at numerator of expression the product that
arises from the difference of 2 squares.

x^2 – y^2 =
(x-y)(x+y)

Putting a^2 + b^2 = x and c^2 = y, we'll
get:

[(a^2+b^2+c^2)*(a^2+b^2-c^2)] = (a^2 + b^2)^2 – c^4
(a^2 + b^2)^2 – c^4 = (1+25)^2 – 1296 = 676-1296 (a^2 + b^2)^2 – c^4  =
-620

We recognize at denominator a perfect square = (a-b)^2
(a-b)^2 = [1 – (-5)]^2 = (1+5)^2 = 36

The value of the expression
is:
(a^2+b^2+c^2)*(a^2+b^2-c^2)]/(a^2+b^2-2ab)=-620/36=-17.22
approx.