In this section, we shall study Solution of a Quadratic Equation by Completing the Square,
Consider the following situation:
The product of Sunita’s age (in years) two years ago and her age four years
from now is one more than twice her present age. What is her present age?
To answer this, let her present age (in years) be x. Then the product of her ages
two years ago and four years from now is (x – 2)(x + 4).
Therefore, (x – 2)(x + 4) = 2x + 1
i.e., x2 + 2x – 8 = 2x + 1
i.e., x2 – 9 = 0
So, Sunita’s present age satisfies the quadratic equation x2 – 9 = 0.
We can write this as x2 = 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 the 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 completely 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 x2 + 4x – 5 = 0? We would probably apply factorisation to
do so, unless we realise (somehow!) that x2 + 4x – 5 = (x + 2)2 – 9.
So, solving x2 + 4x – 5 = 0 is equivalent to solving (x + 2)2 – 9 = 0, which we have
seen is very quick to do. In fact, we can convert any quadratic equation to the form
(x + a)2 – b2 = 0 and then we can easily find its roots. Let us see if this is possible.
Look at Fig. 4.2.
In this figure, we can see how x2 + 4x is being converted to (x + 2)2 – 4.
Consider the following situation:
The product of Sunita’s age (in years) two years ago and her age four years
from now is one more than twice her present age. What is her present age?
To answer this, let her present age (in years) be x. Then the product of her ages
two years ago and four years from now is (x – 2)(x + 4).
Therefore, (x – 2)(x + 4) = 2x + 1
i.e., x2 + 2x – 8 = 2x + 1
i.e., x2 – 9 = 0
So, Sunita’s present age satisfies the quadratic equation x2 – 9 = 0.
We can write this as x2 = 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 the 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 completely 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 x2 + 4x – 5 = 0? We would probably apply factorisation to
do so, unless we realise (somehow!) that x2 + 4x – 5 = (x + 2)2 – 9.
So, solving x2 + 4x – 5 = 0 is equivalent to solving (x + 2)2 – 9 = 0, which we have
seen is very quick to do. In fact, we can convert any quadratic equation to the form
(x + a)2 – b2 = 0 and then we can easily find its roots. Let us see if this is possible.
Look at Fig. 4.2.
In this figure, we can see how x2 + 4x is being converted to (x + 2)2 – 4.