Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
squareroot both sides first
OpenStudy (anonymous):
\[(2y + 3)^2 = 25\]
OpenStudy (anonymous):
\[(2y + 3) = \pm \sqrt{25}\]
OpenStudy (anonymous):
\[(2y+3) = \pm 5\]
OpenStudy (anonymous):
Good.. Go ahead, so what are the two values y could take?
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (lgbasallote):
then subtract 3 from both sides
OpenStudy (anonymous):
I mean this is a quadratic so we should have two answers for y
OpenStudy (anonymous):
squareroot both sides?
OpenStudy (anonymous):
You don't know what square root means?
OpenStudy (anonymous):
I know.
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
but, wait wait.
OpenStudy (anonymous):
ok, then, 5(2y + 3) = 0 ?
OpenStudy (anonymous):
No you are trying to solve for y. which means you are trying to isolate it. Make y be by itself.
Also that sign next to the 5 is a plus or minus. Which means that 5 is both a negative and positive. You can rewrite in the for of
\[2y + 3=\pm 5\]
rewrite it into it's two separate equations one positive and one negative
\[2y + 3=+5\]
\[2y + 3=-5\]
Now solve for y for those two equations.