Question

The number x is 4 more than the number y and their product is 21. Find the sum of the squares of two numbers.

Answer 1

Given:

• The number x is 4 more than the number y, i.e. x = y + 4.
• Their product is 21, i.e. xy = 21.

Step-wise calculation:

1. Substitute x = y + 4 into the product equation:

(y + 4) × y = 21 

y² + 4y = 21

2. Rearranging:

y² + 4y – 21 = 0

3. Solve the quadratic equation:

`y = (-4 +- sqrt(4^2 - 4 xx 1 xx (-21)))/(2 xx 1)`

`y = (-4 +- sqrt(16 + 84))/2`

`y = (-4 +- sqrt(100))/2`

\[ y = \frac{-4 \pm 10}{2} \]

So, `y = (-4 + 10)/2 = 3` or `y = (-4 - 10)/2 = -7`.

4. Corresponding x values:

If y = 3, then x = 3 + 4 = 7.

If y = –7, then x = –7 + 4 = –3.

5. Sum of squares for both cases:

For x = 7, y = 3: 

x² + y² = 7² + 3²

x² + y² = 49 + 9

x² + y² = 58

For x = –3, y = –7: 

(–3)² + (–7)² = 9 + 49

(–3)² + (–7)² = 58 

The sum of the squares of the two numbers is 58.