How do you solve #2b^2 = 162#?

3 Answers
Jul 27, 2018

#b=9 or b=-9#

Explanation:

Here,

#2b^2=162#

Dividing both sides by #2#

#(2b^2)/2=162/2#

#:.b^2=81#

#:.b^2=9^2#

#:.b^2-9^2=0#

#:.(b-9)(b+9)=0to[because x^2-y^2=(x-y)(x+y)]#

#:.b-9=0 or b+9=0#

#:.b=9 or b=-9#

Jul 27, 2018

#b=+-9#

Explanation:

#"divide both sides by 2"#

#b^2=162/2=81#

#color(blue)"take the square root of both sides"#

#b=+-sqrt81larrcolor(blue)"note plus or minus"#

#rArrb=+-9#

Jul 28, 2018

#b=+-9#

Explanation:

We want to isolate #b#. Right now it is being multiplied by #2#, so we can do the inverse, divide by #2# to get

#b^2=81#

Next, we can take the inverse of squaring, which is the square root (of both sides) to get

#b=+-9#

Hope this helps!