[tex]a + 8 = - \dfrac 5 a [/tex]
If I was solving this, which I suppose I am, my first step would be the clear the fractions by multiplying both sides by a.
[tex]a^2 + 8a = - 5[/tex]
That's there, first choice, SELECT IT
Then I'd bring the 5 to the other side;
[tex]a^2 + 8a + 5 = 0[/tex]
Not among the choices.
Then I'd attempt to factor, looking for a pair that multiplies to 5 and adds to 8; no joy there.
Shakespeare Quadratic Formula (2b or -2b) trick: [tex]x^2 - 2bx + c \textrm{ has zeros at } x=b\pm \sqrt{b^2-c}[/tex]
[tex]a = 4 \pm \sqrt{4^2-5} = 4\pm \sqrt{11}[/tex]
All the other choices have nice rational answers so can't be right.
First choice only
