Ask your own question, for FREE!
Mathematics 12 Online
OpenStudy (anonymous):

(2y + 3)^2 = 25

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?

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.

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.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!