7.to.the.power.of.5q..plus.1.equals..49.to.the.power.of.negative.q.squared

Question

It must end up in a quadratic way mostly using the quadratic formula to solve the equation.

Answer 1

7^5q+1 = (49)^ -q^2

7^5q+1 = (7^2)^ -q^2

7^5q+1  = 7^ -2q^2

5q + 1 = - 2q^2               [ if bases are equal than exponants are equal]

2q^2 + 5q + 1 = 0

compare with ax^2 + bx + c = 0

a = 2 ; b = 5 ; c = 1

discriminant = sq.root of b^2 - 4ac

                        = 5^2 - 4*2*1

                        = sq. root of  25 -8

                        = sq.root of 17

q = (- b + sq.roor D)/2a     or q = (- b  - sq.roor D)/2a

q = (-5 + sq.root of 17)/4  or  q = (-5 - sq.root of 17)/4