How do you solve #4b ^ { 2} - 4b - 15= 0#?

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Answer 1 · 2017-05-12T03:00:56

See a solution process below:

First, factor the quadratic on the left side of the equation as:

#(2b + 3)(2b - 5) = 0#

Now, equate each term on the left side of the equation to #0# and solve for #b#:

Solution 1)

#2b + 3 = 0#

#2b + 3 - color(red)(3) = 0 - color(red)(3)#

#2b + 0 = -3#

#2b = -3#

#(2b)/color(red)(2) = -3/color(red)(2)#

#(color(red)(cancel(color(black)(2)))b)/cancel(color(red)(2)) = -3/2#

#b = -3/2#

Solution 2)

#2b - 5 = 0#

#2b - 5 + color(red)(5) = 0 + color(red)(5)#

#2b - 0 = 5#

#2b = 5#

#(2b)/color(red)(2) = 5/color(red)(2)#

#(color(red)(cancel(color(black)(2)))b)/cancel(color(red)(2)) = 5/2#

#b = 5/2#

The solution is: #b = -3/2# and #b = 5/2#