How do I solve these using the square root property? x^2+4x+4=9 (x-5)^2=25
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Originally Posted by mjfaris (x-5)^2=25 This one is pretty easy, taking the square root of each side gives $\displaystyle (x-5)^2=25$ $\displaystyle x-5=\pm 5$ $\displaystyle x=5\pm 5$ $\displaystyle x=0,10$
Originally Posted by mjfaris How do I solve these using the square root property? x^2+4x+4=9 (x-5)^2=25 HI $\displaystyle (x+2)^2=9$ $\displaystyle x+2=\pm 3$ $\displaystyle x=-2\pm 3$ $\displaystyle (x-5)^2=25$ $\displaystyle x-5=\pm 25$ $\displaystyle x=5\pm 25 $
Originally Posted by mathaddict HI $\displaystyle (x+2)^2=9$ $\displaystyle x+2=\pm 3$ $\displaystyle x=-2\pm 3$ $\displaystyle (x-5)^2=25$ $\displaystyle x-5=\pm 25$ $\displaystyle x=5\pm 25 $ You didn't square root 25.
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