# 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 equation.

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### 1 Answer

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.**