• October 26th 2008, 02:55 PM
Gigabytes
What are the zeros of y-7=4x^2

and Solve square root of 2x^2 - sqaure root of 2 = 0
• October 26th 2008, 04:38 PM
terr13
$y-7 = 4x^2$
$-7 = 4x^2$
Never true in the reals.
$\sqrt(2x^2) - \sqrt(2) =0$
$(\sqrt(2x^2)-\sqrt(2))(\sqrt(2x^2)+\sqrt(2)) =0$
$2x^2 -2 =0$
$x^2 -1 = 0$
$x = -1 \, or\, 1$