# Solving quadratic equations/square root property

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• December 2nd 2009, 07:47 PM
mjfaris
Solving quadratic equations/square root property
How do I solve these using the square root property?

x^2+4x+4=9

(x-5)^2=25
• December 2nd 2009, 07:57 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, 08:01 PM
mathaddict
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, 09:53 AM
alibond07
Quote:

Originally Posted by mathaddict
HI

$(x+2)^2=9$

$x+2=\pm 3$

$x=-2\pm 3$

$(x-5)^2=25$

$x-5=\pm 25$

$
x=5\pm 25
$

You didn't square root 25.