Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

x^{2} + 6x – (a^{2} + b^{2} – 8) = 0

Given: x^{2} + 6x – (a^{2} + b^{2} – 8) = 0

Comparing with standard quadratic equation Ax^{2} + Bx + C = 0

A = 1, B = 6, C = – (a^{2} + b^{2} – 8)

Discriminant D = B^{2} – 4AC

= (6)^{2} – 4.1. – (a^{2} + b^{2} – 8)

= 36+ 4a^{2} + 8a – 32 = 4a^{2} + 8a + 4

= 4(a^{2} + 2a + 1)

= 4(a + 1)^{2} > 0 Using a^{2} + 2ab + b^{2} = (a + b)^{2}

Hence the roots of equation are real.

= 2(a + 1)

Roots are given by

x = (a – 2) or x = – (4 + a)

Hence the roots of equation are (a – 2) or – (4 + a)

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