Courtney T.
asked • 01/16/13Find all real number solutions for 4y^2-25=0
I got it to zero but im stuck on what to do after that
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1 Expert Answer
Robert J. answered • 01/16/13
Tutor
4.6
(13)
Certified High School AP Calculus and Physics Teacher
Method I.
Factor out,
(2y+5)(2y-5) = 0
2y+5 = 0 => y = -5/2
2y-5 = 0 => y = 5/2
Method II.
Isolate y^2,
y^2 = 25/4
Take square roots on both sides,
y = +/- 5/2
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Herb K.
4y^2 - 25 = 0; add 25 to each side of equation ---> 4y^2 = 25; now, divide each side fo equation by 4 ---> y^2 = 25/4; now, use square root property (i.e., take square roots) ---> y = (+/-)(5/2)
01/17/13