Solution of basic linear equation, Algebra

Solve following equations.

(a)   x2 -100 = 0

 (b) 25 y 2 - 3 = 0


There actually isn't all that much to these problems.  To use the square root property all that we have to do is get the squared quantity on the left side by itself along with a coefficient of 1 and the number on the other side. Once it is done we can employ the square root property.


             x2 -100 + 0

It is a fairly simple problem thus here is the work for this equation.

x2  = 100

 x = ± 100 = ±10

Thus, there are two solutions to this equation,

x =±10 .  Remember it means that there are really two solutions here, x = -10 and x =10 .

 (b) 25 y 2 - 3 = 0

The main difference among this one & the previous one is the 25 in front of the squared term.  The square root property desired a coefficient of one there. That's simple enough to deal with but; we'll just divide both sides through 25.  Here is the work for this equation.

                                          25 y 2  = 3

y 2  = 3/25

⇒         y = ±√(3/5) = ±(√3 )/5

In this the solutions are a little messy, but several of these will do so don't worry regarding that. Also note that since we knew what the square root of 25 was we went ahead & split the square root of the fraction up as illustrated.  Again, recall that there are actually two solutions here, one positive & one negative.

Posted Date: 4/6/2013 4:04:27 AM | Location : United States

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