Use square roots to solve the following equations ; round to the nearest hundredth? -2w2+ 201.02 =66.54. Second problem is 3y2+51=918?

1 Answer
Ben
Mar 21, 2018
  1. #w=+-8.2#
  2. #y=+-17#

Explanation:

I'm going to make an assumption that the equations look like this:

  1. #-2w^2+201.02=66.54#
  2. #3y^2+51=918#

Let's solve the first problem:

First, move the additive term to the right hand side:
#-2w^2cancel(+201.02-201.02)=66.54-201.02#
#-2w^2=-134.48#

Next, divide by any constant coefficients:
#(-2w^2)/(-2)=(-134.48)/(-2) rArr w^2=67.24#

Finally, take the square root from both sides. Remember, any real number squared comes out positive, so the root of a given number can be both positive and negative:

#sqrt(w^2)=sqrt(67.24)#

#color(red)(w=+-8.2)#

Now, we'll do problem 2 using the same steps:

#3y^2cancel(+51-51)=918-51 rArr 3y^2=867#

#(3y^2)/3=867/3 rArr y^2=289#

#sqrt(y^2)=sqrt(289)#

#color(blue)(y=+-17)#

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