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

  • $\begingroup$ Are you trying to perform the multiplication or solve the equation for b? $\endgroup$ – NoChance Oct 3 '15 at 19:05
  • $\begingroup$ I am trying to perform multiplication and ultimately try to solve the equation.. $\endgroup$ – MATH ASKER Oct 3 '15 at 19:08
  • $\begingroup$ a bracket is missing $\endgroup$ – 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 $$\begin{align}(-\sqrt{7 - 2b} + -2)^2 &= (-\sqrt{7 - 2b})^2 + 2(-\sqrt{7 - 2b})(-2) + (-2)^2\\&=(7 - 2b) + 4\sqrt{7 - 2b} + 4\\&=11 - 2b + 4\sqrt{7 - 2b} \end{align}$$

Which is as far as you can take it.

  • $\begingroup$ 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 $${(a+b)^2}$$ $\endgroup$ – MATH ASKER Oct 3 '15 at 20:03
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    $\begingroup$ I suspect what your teacher was after was to get you to practice multiplying out expressions, as I did to derive the formula, so that you would come to understand why the formula is true. This would be far more helpful to you in the long run than memorizing and using formulas that you don't understand. I never memorized that formula on purpose (though I've used it so many times I know it anyway). Because I know how to distribute, I never needed to memorize it or "the foil method" or any of dozens of other tricks that some people have to learn. $\endgroup$ – Paul Sinclair Oct 3 '15 at 20:13

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