The sum of the squares of two consecutive multiples of 7 is 1225. Find the multiples.

Answer

To solve the problem of finding two consecutive multiples of 7 whose squares sum up to 1225, we will follow these steps:

### Step 1: Define the Variables
Let the first multiple of 7 be \( 7x \). Since we are looking for consecutive multiples, the next consecutive multiple will be \( 7(x + 1) \).

### Step 2: Set Up the Equation
According to the problem, the sum of the squares of these two multiples is 1225. Therefore, we can write the equation:
\[ (7x)^2 + (7(x + 1))^2 = 1225 \]

### Step 3: Expand the Squares
Now, we will expand the squares in the equation:
\[ (7x)^2 = 49x^2 \]
\[ (7(x + 1))^2 = 49(x + 1)^2 = 49(x^2 + 2x + 1) = 49x^2 + 98x + 49 \]
So, substituting back into the equation, we have:
\[ 49x^2 + (49x^2 + 98x + 49) = 1225 \]

### Step 4: Combine Like Terms
Now, combine the like terms:
\[ 49x^2 + 49x^2 + 98x + 49 = 1225 \]
This simplifies to:
\[ 98x^2 + 98x + 49 = 1225 \]

### Step 5: Rearrange the Equation
Next, we will move 1225 to the left side of the equation:
\[ 98x^2 + 98x + 49 - 1225 = 0 \]
This simplifies to:
\[ 98x^2 + 98x - 1176 = 0 \]

### Step 6: Simplify the Equation
We can divide the entire equation by 2 to make it simpler:
\[ 49x^2 + 49x - 588 = 0 \]

### Step 7: Factor the Quadratic Equation
Now, we will factor the quadratic equation. We need two numbers that multiply to \( 49 \times -588 = -28812 \) and add to \( 49 \). The numbers are \( 84 \) and \( -35 \):
\[ 49x^2 + 84x - 35x - 588 = 0 \]
Grouping the terms:
\[ (49x^2 + 84x) + (-35x - 588) = 0 \]
Factoring by grouping:
\[ 7x(7x + 12) - 49(7x + 12) = 0 \]
Factoring out \( (7x + 12) \):
\[ (7x + 12)(7x - 49) = 0 \]

### Step 8: Solve for \( x \)
Setting each factor to zero gives us:
\[ 7x + 12 = 0 \quad \Rightarrow \quad x = -\frac{12}{7} \quad \text{(not valid)} \]
\[ 7x - 49 = 0 \quad \Rightarrow \quad x = 7 \]

### Step 9: Find the Multiples
Now, substituting \( x = 7 \) back into our expressions for the multiples:
- First multiple: \( 7x = 7 \times 7 = 49 \)
- Second multiple: \( 7(x + 1) = 7 \times (7 + 1) = 7 \times 8 = 56 \)

### Conclusion
The two consecutive multiples of 7 are **49 and 56**.

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