Quadratic Problem solving question

Question

The sum of the length and breadth of a rectangle is 15cm. Find the sides if the area is 56cm

Answer 1

Since the sum of length (Let it = L) and breadth (Let it = B) is 15cm, and the area (Let it = A) is 56cm^2 which, for a rectangle, is calculated as L x B:

A = LB and L+ B = 15 so, subtracting B from both sides L = 15 - B

then, by substitution A = (15 - B) B = 15B - B^2 = 56

subtracting 56 from both sides

we have the quadratic: 15B - B^2 - 56 = 0

Multiplying through by - 1 and rearranging into standard form gives:

B^2 - 15B + 56 = 0

From the + on the 56 we know that it is made up of two negative factors or two positive factors that add to the middle coefficient of -15, thus both are negative. Going through factors of 56 the factors 2 and 28, 4 and 14, do not add to 15. the factors -7 and -8 multiply to +56 and add to -15, so

B^2 - 15B + 56 = (B-7) (B-6) = 0

Since one or both of the binomial factors must = 0 for the multiplication to = 0

B - 7 = 0 and/or B - 8 = 0

and B = 7 or B = 8

Going back to L + B = 15 either B = 7 and L = 8 or L = 7 and B = 8

Either satisfies the problem since no further information is given - however, if your class always uses length as the longest side of a rectangle, or always uses breadth as the longest side, this could make one answer more correct than the other.

Answer 2

The answer is clearly 7 and 8; however, working it out, we have

l+w = 15

lw (area) = 56

We can solve by l in the second equation by substituting in the first w = 15 - l

Then, l(15-l) = 56

15l - l^2 = 56

l^2 - 15l + 56 = 0

Note that we are actually right back where we started.

This factors as (l - 8) (l - 7) = 0

l = 7, 8; of course, as l+w = 15, this just gets us back to l = 8, w = 7 (or the opposite unless you want l to be greater than w).