# Solving quadratic equations/square root property

• December 2nd 2009, 06:47 PM
mjfaris
How do I solve these using the square root property?

x^2+4x+4=9

(x-5)^2=25
• December 2nd 2009, 06:57 PM
pickslides
Quote:

Originally Posted by mjfaris

(x-5)^2=25

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

$(x-5)^2=25$

$x-5=\pm 5$

$x=5\pm 5$

$x=0,10$
• December 2nd 2009, 07:01 PM
Quote:

Originally Posted by mjfaris
How do I solve these using the square root property?

x^2+4x+4=9

(x-5)^2=25

HI

$(x+2)^2=9$

$x+2=\pm 3$

$x=-2\pm 3$

$(x-5)^2=25$

$x-5=\pm 25$

$
x=5\pm 25
$
• December 3rd 2009, 08:53 AM
alibond07
Quote:

HI

$(x+2)^2=9$

$x+2=\pm 3$

$x=-2\pm 3$

$(x-5)^2=25$

$x-5=\pm 25$

$
x=5\pm 25
$

You didn't square root 25.