How do you solve #4b ^ { 2} - 25= 0#?

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Answer 1 · 2018-05-31T20:25:06

#b = 5/2#

#4b^2 - 25 = 0#

Adding #25# to both sides;

#4b^2 - 25 + 25 = 0 + 25#

#4b^2 + 0 = 0 + 25#

#4b^2 = 25#

Dividing both sides by the coefficient of #b^2;#

#(4b^2)/4 = 25/4#

#(cancel4b^2)/cancel4 = 25/4#

#b^2 = 25/4#

Square root both sides;

#sqrt(b^2) = sqrt25/4#

#b = 5/2#