Question

Which equation has both 3 and -3 as possible values of y ?

Answer 1

Answer:

${y}^{2} - 9 = 0$

Step-by-step explanation:

as both 3 and - 3 are possible values of y;

$y = 3 \: or \: y = - 3$

$y - 3 = 0 \: or \: y + 3 = 0$

therefore;

$(y - 3)(y + 3) = 0$

by expanding;

${y}^{2} - {3}^{2} = 0$

${y }^{2} - 9 = 0$