The product of Sunita’s age (in years) two years ago and her age four years hence is one
more than twice her present age. What is her present age?
To answer this, let her present age be x years. Their age two years ago would be x – 2 and
the age after four years will be x + 4. So, the product of both the ages is (x – 2)(x + 4).
Therefore, (x – 2)(x + 4) = 2x + 1
i.e., x
2
+ 2x – 8 = 2x + 1
i.e., x
2
– 9 = 0
So, Sunita’s present age satisfies the quadratic equation x
2
– 9 = 0.
We can write this as x
2
= 9. Taking square roots, we get x = 3 or x = – 3. Since the age is
a positive number, x = 3.
So, Sunita’s present age is 3 years.
Now consider another quadratic equation (x + 2)2
– 9 = 0. To solve it, we can write it as
(x + 2)2
= 9. Taking square roots, we get x + 2 = 3 or x + 2 = – 3.
Therefore, x = 1 or x = –5
So, the roots of the equation (x + 2)2
– 9 = 0 are 1 and – 5.
In both the examples above, the term containing x is inside a square, and we found the
roots easily by taking the square roots. But, what happens if we are asked to solve the equation
x
2
+ 4x – 4 = 0, which cannot be solved by factorisation also.
So, we now introduce the method of completing the square. The idea behind this
method is to adjust the left side of the quadratic equation so that it becomes a perfect
square of the first degree polynomial and the RHS without x term.
The process is as follows:
x
2
+ 4x – 4 = 0
Þ x
2
+ 4x = 4
x
2
+ 2. x . 2 = 4
Now, the LHS is in the form of a
2
+ 2ab. If we add b
2
it becomes as a
2 + 2ab + b
2
which
is perfect square. So, by adding b
2
= 22
= 4 to both sides we get,