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Find 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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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)

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1 Answer

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