Solving quadratic equations/square root property

**mjfaris** · December 2nd 2009, 06:47 PM

How do I solve these using the square root property?

x^2+4x+4=9

(x-5)^2=25

**pickslides** · December 2nd 2009, 06:57 PM

Originally Posted by mjfaris

(x-5)^2=25

This one is pretty easy, taking the square root of each side gives

$(x-5)^2=25$

$x-5=\pm 5$

$x=5\pm 5$

$x=0,10$

**mathaddict** · December 2nd 2009, 07:01 PM

Originally Posted by mjfaris

How do I solve these using the square root property?

x^2+4x+4=9

(x-5)^2=25

HI

$(x+2)^2=9$

$x+2=\pm 3$

$x=-2\pm 3$

$(x-5)^2=25$

$x-5=\pm 25$

$<br /> x=5\pm 25 <br />$

**alibond07** · December 3rd 2009, 08:53 AM

Originally Posted by mathaddict

HI

$(x+2)^2=9$

$x+2=\pm 3$

$x=-2\pm 3$

$(x-5)^2=25$

$x-5=\pm 25$

$<br /> x=5\pm 25 <br />$

You didn't square root 25.

Thread: Solving quadratic equations/square root property