Quadratic Equation Example

Q: Sum of the area of two squares is 400 square cm. If the difference of their perimeters is 16 cm, find the sides of the two squares.

                                                                  SOLUTION

Let the side of the first square be ‘a’ and that of the second square be ‘b’.  then, according to the problem,

                                         a^2 + b^2 = 400   …(1)

and                                  4a – 4b = 16          …(2)    Hint : Perimeter = 4 times the side of a square

Now, from (2), a = b + 4,

substituting this value of a in (1), we have

                             (4+b) ^2+b^2 = 400    

Expanding the above we get the following quadratic equation:

                                                b^2 + 4*b - 192 = 0             ….(3)

factoring the polynomial, we get,

                                               b^2 + 16b-12b-192 = 0        …. (4)

simplifying, we obtain,

                                              b(b+ 16)-12(b+ 16) = 0

• (b-12) (b+16) = 0

From which we get two solutions viz

                                             b = 12 or b = -16

Now, since the side of a square cannot be negative, thus the correct solution is b = 12, whence, a = b + 4.

Thus, a = 16.

Therefore, the sides of the square are 12 and 16 cm respectively.