what are the steps in solving for w, if w(w+2)=15

Question

If the length of rectangle is 2 units more than it's length, and it's area is 15 square units, what are the rectangle's dimensions?

Answer 1

This is a quadratic equation problem. Apply the distributive law on the left side of the equation in which you get:

 

w^2 + 2w=15

 

Subtract 15 from both sides of the equation in which is:

 

w^2 + 2w - 15 = 0

 

The way to solve this equation is to use the foil method in which you could factor the left side of the equation:

 

(w + 5)(w - 3) = 0

 

w = 3 units Note: You could see the width of a rectangle can't have a negative measurement.

 

This implies l = 3+2=5 units

 

Therefore, the dimensions of the rectangle are 3 and 5 units.

Answer 2

This particular equation is a quadratic, and so to solve it you will need to expand the left side and set it equal to zero:

 

w^2+2w=15

 

w^2+2w-15=0

 

At this point you your quadratic is in standard form and is ready to be solved like any other quadratic.  This one factors rather easily:

 

(w+5)(w-3)=0

 

So the two solutions here are w=-5,3.  However, since the width of a rectangle can't be negative, we can eliminate -5 as an answer.  Thus, the width is 3.  You can then find the length by plugging this width into wl=15.