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.
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.