Answer
(a) By factoring the expression we got:
$y^4+2y^2-3=(y^2-1)(y^2+3)$
(b) The two solutions are:
$y_{1}=\pm1$
Work Step by Step
$y^4+2y^2-3$
Let $y^2=z$
Then we have:
$y^4+2y^2-3=z^2+2z-3=z^2+3z-z-3=z(z+3)-1(z+3)=(z-1)(z+3)$
(a) By factoring the expression we got:
$(y^2-1)(y^2+3)$
(b) $(y^2-1)(y^2+3)=0$
The two solutions possible are:
$y_{1}^2=1$
or
$y_{2}^2=-3$
No square number can ba negative, therefore $y_{2}$ is not a possible solution.
The two other solutions are:
$y_{1}=\pm1$
