That's good, subbing y for m^2. Therefore, you get a quadratic.
Just factor it.
Now, sub m^2 back in:
Now, The M^2+6 results in 2 non-real solutions.
The 5m^2-6 will give two real solutions.
Hello, OzzMan!
Solve for
Let
Then i plugged it into the quadratic formula:
From there i did
which gives an imaginary answer.
P.S. I used as my answer and I got marked off 1 point.
Anyone know why?
Just a guess . . . maybe you were expected to rationalize that square root?
. .
Oh yes, your reasoning and your work is correct . . . Nice going!