How do you solve #\sqrt { 10- 3b } = b#?

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Answer 1 · 2017-12-19T17:49:40

b=2

#sqrt(10-3b)=b#

Just square bost side os the equality:

#(10-3b)=b^2#

#0=b^2+3b-10#

#b=(-3+-sqrt(3^2-4*1*(-10)))/2#

#b=(-3+-sqrt(49))/2#

#b=(-3+-7)/2#

#b=(-3+7)/2# ou #b=(-3-7)/2#

#b=2# ou #b=-5#

You must determine the dominion of this expression:

since there is a square root and the solution must be a posiive number

#10-3b>0# and #b>0#

#10>3b#

#b<10/3# and #b>0#

Concernig this, only b=2 is valid