14.

The difference of two numbers is 4, and the sum of their squares is 40.

a)
(i) If the greater number is [tex]\( x \)[/tex], write the smaller number in terms of [tex]\( x \)[/tex].
(ii) Find the numbers.
(iii) Find the ratio of the sum to the product of the numbers.

b) The sum of the squares of two consecutive natural even numbers is 100.



Answer :

Certainly! Let's go through the problem step by step.

### Problem Details:
1. The difference between two numbers is 4.
2. The sum of their squares is 40.

### Steps to Solve:

(i) Express the smaller number in terms of the greater number:

Let the greater number be [tex]\( x \)[/tex].
Since the difference between the two numbers is 4, the smaller number will be [tex]\( x - 4 \)[/tex].

(ii) Find the numbers:

Given the sum of their squares is 40, we can write the equation:
[tex]\[ x^2 + (x - 4)^2 = 40 \][/tex]

Expanding and simplifying the equation:
[tex]\[ x^2 + (x^2 - 8x + 16) = 40 \][/tex]
[tex]\[ 2x^2 - 8x + 16 = 40 \][/tex]
[tex]\[ 2x^2 - 8x + 16 - 40 = 0 \][/tex]
[tex]\[ 2x^2 - 8x - 24 = 0 \][/tex]
[tex]\[ x^2 - 4x - 12 = 0 \][/tex]

We solve the quadratic equation [tex]\( x^2 - 4x - 12 = 0 \)[/tex] using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -4 \)[/tex], and [tex]\( c = -12 \)[/tex]:

[tex]\[ x = \frac{4 \pm \sqrt{16 + 48}}{2} \][/tex]
[tex]\[ x = \frac{4 \pm \sqrt{64}}{2} \][/tex]
[tex]\[ x = \frac{4 \pm 8}{2} \][/tex]

Thus, we have two possible solutions for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{12}{2} = 6 \][/tex]
[tex]\[ x = \frac{-4}{2} = -2 \][/tex]

Now, let's find the corresponding smaller numbers:
If [tex]\( x = 6 \)[/tex], then the smaller number is [tex]\( 6 - 4 = 2 \)[/tex].
If [tex]\( x = -2 \)[/tex], then the smaller number is [tex]\( -2 - 4 = -6 \)[/tex].

So, the numbers are:
1. 6 and 2
2. -2 and -6

(iii) Find the ratio of the sum to the product of the numbers:

We need to find the ratio of the sum of these numbers to their product.

For the first pair (6 and 2):
- Sum: [tex]\( 6 + 2 = 8 \)[/tex]
- Product: [tex]\( 6 \cdot 2 = 12 \)[/tex]
- Ratio: [tex]\( \frac{8}{12} = \frac{2}{3} \)[/tex]

For the second pair (-2 and -6):
- Sum: [tex]\( -2 + (-6) = -8 \)[/tex]
- Product: [tex]\( -2 \cdot -6 = 12 \)[/tex]
- Ratio: [tex]\( \frac{-8}{12} = \frac{-2}{3} \)[/tex]

Thus, the ratio of the sum to the product of the numbers can be either [tex]\( \frac{2}{3} \)[/tex] for the first pair or [tex]\( \frac{-2}{3} \)[/tex] for the second pair.