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.
