The problem is $$0 = -(\sqrt{7 -2 b} - 2) * -\sqrt{7-2 b} - 2)$$

the original question was this $$-2 = \sqrt{7 -2 b} - \sqrt{2 b + 3}$$ and than i found out the you had to isolate one radical version so i substracted $$-(\sqrt{7 -2 b} - 2)$$ and than you get $$(-\sqrt{7 -2 b} -2)^2 = (-\sqrt{2 b +3})^2$$ i got stuck after trying to square the side with $$(-\sqrt{7 -2 b} -2)^2$$ can anyone tell me how you're supposed to square/multiply them? I am trying to perform multiplication and ultimately try to solve the equation...if anyone knows how to solve the equation the way i have shown above please help me as i have been stuck with this equation for long time

• Are you trying to perform the multiplication or solve the equation for b? – NoChance Oct 3 '15 at 19:05
• I am trying to perform multiplication and ultimately try to solve the equation.. – MATH ASKER Oct 3 '15 at 19:08
• a bracket is missing – Dr. Sonnhard Graubner Oct 3 '15 at 19:23

\begin{align}(x + y)^2 &= (x + y)(x+y) \\ &= x(x +y) + y(x +y) \\ &= (x^2 + xy) + (xy + y^2) \\ &= x^2 + 2xy + y^2\end{align}
Now in your case $x = -\sqrt{7 - 2b}$ and $y = -2$, so \begin{align}(-\sqrt{7 - 2b} + -2)^2 &= (-\sqrt{7 - 2b})^2 + 2(-\sqrt{7 - 2b})(-2) + (-2)^2\\&=(7 - 2b) + 4\sqrt{7 - 2b} + 4\\&=11 - 2b + 4\sqrt{7 - 2b} \end{align}
• Thanks a lot for answering...i remember using this a^2 +2ab + b^2 FORMAULA so much when i was in asia but when i moved to the west...the teacher had told me to not use this formula...so does that mean i can use this formula if the problem is $${(a+b)^2}$$ – MATH ASKER Oct 3 '15 at 20:03