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Solve for bb.\newline2b+2=4b\sqrt{2b + 2} = \sqrt{4b}\newlineb=b = _____

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Q. Solve for bb.\newline2b+2=4b\sqrt{2b + 2} = \sqrt{4b}\newlineb=b = _____
  1. Square Both Sides: Square both sides of the equation to eliminate the square roots.\newline(2b+2)2=(4b)2(\sqrt{2b + 2})^2 = (\sqrt{4b})^2\newline2b+2=4b2b + 2 = 4b
  2. Subtract and Simplify: Subtract 2b2b from both sides to get the bb terms on one side and the constants on the other side.\newline2b+22b=4b2b2b + 2 - 2b = 4b - 2b\newline2=2b2 = 2b
  3. Divide to Solve: Divide both sides by 22 to solve for bb.22=2b2\frac{2}{2} = \frac{2b}{2}1=b1 = b

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