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How do you solve $3 = \sqrt{b - 1}$ $3=sqrt(b-1)$ ?

1 Answer

MathFact-orials.blogspot.com

Sep 19, 2016

$b = 10$ $b=10$

Explanation:

Starting with:

$3 = \sqrt{b - 1}$ $3=sqrt(b-1)$

we can square both sides to get:

$3^{2} = {\sqrt{b - 1}}^{2}$ $3^2=sqrt(b-1)^2$

$9 = b - 1$ $9=b-1$

$b = 10$ $b=10$

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