• March 23rd 2011, 10:56 AM
AlyoshaKaz
Can anyone help me understand how we get from

$\frac{1}{8}(56 \pm \sqrt{64 +16y})$

to

$7 \pm \sqrt{1 + \frac{1}{4}y}$

thanks
• March 23rd 2011, 10:59 AM
mathfun
$\tfrac{1}{8}\left( {56 \pm \sqrt {64 + 16y} } \right) = \tfrac{{56}}{8} \pm \tfrac{{\sqrt {64 + 16y} }}{8} = 7 \pm \tfrac{{\sqrt {64} \sqrt {1 + \tfrac{1}{4}y} }}{8} = 7 \pm \sqrt {1 + \tfrac{1}{4}y}$
• March 23rd 2011, 11:17 AM
AlyoshaKaz
so like

$\sqrt{64+16y} = \sqrt{64(1 + \frac{1}{4}y)} = \sqrt{64} \sqrt{1 + \frac{1}{4}y}$

?
• March 23rd 2011, 11:19 AM
mathfun
yeap.
Just keep in mind that $\sqrt {a \cdot b} = \sqrt a \cdot \sqrt b$