Let's 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., `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 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 `x^2 + 4x – 5 = 0` We would probably apply factorisation to do so, unless we realise (somehow!) that `x^2 + 4x – 5 = (x + 2)^2 – 9.`
So, solving `x^2 + 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 – b^2 = 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 `x^2 + 4x` is being converted to` (x + 2)^2 – 4.`
The process is as follows:
`x^2+4x = ( x^2+4/2 x ) +4/2 x`
`= x^2 + 2x + 2x`
`= (x + 2) x + 2 × x`
`= (x + 2) x + 2 × x + 2 × 2 – 2 × 2`
`= (x + 2) x + (x + 2) × 2 – 2 × 2`
`= (x + 2) (x + 2) – 2^2`
`= (x + 2)^2 – 4`
So, `x^2 + 4x – 5 = (x + 2)^2 – 4 – 5 = (x + 2)^2 – 9`
So, `x^2 + 4x – 5 = 0` can be written as `(x + 2)^2 – 9 = 0` by this process of completing the square. This is known as the method of completing the square.
In brief, this can be shown as follows:
`x^2+4x= (x+4/2)^2 -(4/2)^2 = (x+4/2)^2-4`
So, `x^2 + 4x – 5 = 0` can be rewritten as
`(x+4/2)^2-4-5 = 0`
`(x+2)^2-9 = 0`
Consider now the equation `3x^2 – 5x + 2 = 0.` Note that the coefficient of `x^2` is not
a perfect square. So, we multiply the equation throughout by 3 to get
`9x^2 – 15x + 6 = 0`
Now `9x^2-15x+6 = (3x)^2-2xx3x xx 5/2 +6`
`= (3x)^2-2xx3x xx 5/2 +(5/2)^2 - (5/2)^2 +6`
` = (3x-5/2)^2 -25/4 +6 = ( 3x -5/2)^2 - 1/4`
So, `9x^2 – 15x + 6 = 0` can be written as
`( 3x - 5/2)^2 - 1/4 = 0`
`(3x - 5/2)^2 = 1/4`
So, the solutions of `9x^2 – 15x + 6 = 0` are the same as those of `( 3x - 5/2)^2 = 1/4`.
`3x - 5/2 = 1/2 ` or `3x - 5/2 = -1/2`
(We can also write this as `3x - 5/2 = pm 1/2` , where ‘±’ denotes ‘plus minus’.)
Thus `3x= 5/2+1/2 ` or `3x = 5/2-1/2`
So `x = 5/6 +1/6` or `x = 5/6 - 1/6`
Therefore `x = 1` or `x = 4/6`
`x = 1` or `x = 4/6`
i.e. `x = 1` or `x = 2/3`
Therefore, the roots of the given equation are 1 and `2/3`
Remark : Another way of showing this process is as follows :
The equation `3x^2 - 5x+2 = 0`
is the same as `x^2 - 5/3 x +2/3 = 0`
Now `x^2 - 5/3 x +2/3 = { x - 1/2 (5/3)}^2 - { 1/2 (5/3)}^2 +2/3`
` = ( x-5/6)^2 +2/3 - 25/36`
` = ( x-5/6)^2 - 1/36 = ( x-5/6)^2 - ( 1/6)^2`
So, the solutions of `3x^2 – 5x + 2 = 0` are the same as those of `( x-5/6)^2 - (1/6)^2 = 0`
which are `x - 5/6 = pm 1/6`
i.e `x = 5/6 +1/6 = 1` and `x = 5/6-1/6 = 2/3`
Let us consider some examples to illustrate the above process.
Let's 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., `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 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 `x^2 + 4x – 5 = 0` We would probably apply factorisation to do so, unless we realise (somehow!) that `x^2 + 4x – 5 = (x + 2)^2 – 9.`
So, solving `x^2 + 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 – b^2 = 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 `x^2 + 4x` is being converted to` (x + 2)^2 – 4.`
The process is as follows:
`x^2+4x = ( x^2+4/2 x ) +4/2 x`
`= x^2 + 2x + 2x`
`= (x + 2) x + 2 × x`
`= (x + 2) x + 2 × x + 2 × 2 – 2 × 2`
`= (x + 2) x + (x + 2) × 2 – 2 × 2`
`= (x + 2) (x + 2) – 2^2`
`= (x + 2)^2 – 4`
So, `x^2 + 4x – 5 = (x + 2)^2 – 4 – 5 = (x + 2)^2 – 9`
So, `x^2 + 4x – 5 = 0` can be written as `(x + 2)^2 – 9 = 0` by this process of completing the square. This is known as the method of completing the square.
In brief, this can be shown as follows:
`x^2+4x= (x+4/2)^2 -(4/2)^2 = (x+4/2)^2-4`
So, `x^2 + 4x – 5 = 0` can be rewritten as
`(x+4/2)^2-4-5 = 0`
`(x+2)^2-9 = 0`
Consider now the equation `3x^2 – 5x + 2 = 0.` Note that the coefficient of `x^2` is not
a perfect square. So, we multiply the equation throughout by 3 to get
`9x^2 – 15x + 6 = 0`
Now `9x^2-15x+6 = (3x)^2-2xx3x xx 5/2 +6`
`= (3x)^2-2xx3x xx 5/2 +(5/2)^2 - (5/2)^2 +6`
` = (3x-5/2)^2 -25/4 +6 = ( 3x -5/2)^2 - 1/4`
So, `9x^2 – 15x + 6 = 0` can be written as
`( 3x - 5/2)^2 - 1/4 = 0`
`(3x - 5/2)^2 = 1/4`
So, the solutions of `9x^2 – 15x + 6 = 0` are the same as those of `( 3x - 5/2)^2 = 1/4`.
`3x - 5/2 = 1/2 ` or `3x - 5/2 = -1/2`
(We can also write this as `3x - 5/2 = pm 1/2` , where ‘±’ denotes ‘plus minus’.)
Thus `3x= 5/2+1/2 ` or `3x = 5/2-1/2`
So `x = 5/6 +1/6` or `x = 5/6 - 1/6`
Therefore `x = 1` or `x = 4/6`
`x = 1` or `x = 4/6`
i.e. `x = 1` or `x = 2/3`
Therefore, the roots of the given equation are 1 and `2/3`
Remark : Another way of showing this process is as follows :
The equation `3x^2 - 5x+2 = 0`
is the same as `x^2 - 5/3 x +2/3 = 0`
Now `x^2 - 5/3 x +2/3 = { x - 1/2 (5/3)}^2 - { 1/2 (5/3)}^2 +2/3`
` = ( x-5/6)^2 +2/3 - 25/36`
` = ( x-5/6)^2 - 1/36 = ( x-5/6)^2 - ( 1/6)^2`
So, the solutions of `3x^2 – 5x + 2 = 0` are the same as those of `( x-5/6)^2 - (1/6)^2 = 0`
which are `x - 5/6 = pm 1/6`
i.e `x = 5/6 +1/6 = 1` and `x = 5/6-1/6 = 2/3`
Let us consider some examples to illustrate the above process.