This problem I am just carious about not homework?

Question

ok the problem i would like to know how to set up is:

3- a - 5 / a+3 = a^2 - 1 / a + 3

Answer 1

I think multiply the num and denom by a+3 square it

i don't know

Answer 2

3 - (a - 5)/(a + 3) = (a^2 - 1)/(a + 3)

[a + 3][3 - (a - 5)/(a + 3)] = [a + 3][(a^2 - 1)/(a + 3)]

3(a + 3) - a + 5 = a^2 - 1

3a + 9 - a + 5 = a^2 - 1

a^2 - 3a + a - 1 - 9 - 5 = 0

a^2 - 2a - 15 = 0

a^2 + 3a - 5a - 15 = 0

(a^2 + 3a) - (5a + 15) = 0

a(a + 3) - 5(a + 3) = 0

(a + 3)(a - 5) = 0

a + 3 = 0

a = -3

a - 5 = 0

a = 5

∴ a = -3 , 5

Answer 3

(3-a-5)/(a+3) = (a^2-1)/a+3

a+3 gets cancelled on both the sides

3-a-5 = a^2-1

a^2 + a +1 =0

using the formula [-b+/- sqrt(b^2-4ac)]/2a

             where a=1,b=1,c=1

        a=(-1+ i sqrt3)/2,(-1- i sqrt3)/2

                              

Answer 4

(3 - a - 5) /(a + 3) = (a^2 - 1)/(a + 3)

=> (- a - 2)/(a + 3) = (a^2 - 1)/(a +3)

=>- a - 2 = a^2 - 1 [taking a # - 3]

=> a^2 + a +1 = 0

=> a = { -1 +/- sqrt[1 - 4.1.1]}/2

=> a = { -1 +/- sqrt(1 - 4)}/2

=> a = { - 1 +/- sqrt(-3)}/2

=> a = - 1/2 +/- i.sqrt(3)/2 [where i^2 = - 1 => i = sqrt(i)] <==ANSWER