Question

Let $x$ and $y$ be positive integers such that $x^2 + y^2 + xy = 61$. If $x = 4$, then the value of $y$ is

• A. 4
• B. 5
• C. 6
• D. 7

Hint:

After substitution, simplify to a quadratic and factorize for integer solutions.

Answer 1

The Correct Option is B

To solve the given equation \(x^2 + y^2 + xy = 61\) with \(x = 4\), we need to find the integer value of \(y\).

1. Substitute \(x = 4\) into the equation: \(4^2 + y^2 + 4y = 61\).
2. Calculate \(4^2\): \(16 + y^2 + 4y = 61\).
3. Rearrange the equation to isolate terms: \(y^2 + 4y + 16 = 61\).
4. Subtract 16 from both sides to simplify: \(y^2 + 4y = 45\).
5. Bring all terms to one side to form a quadratic equation: \(y^2 + 4y - 45 = 0\).
6. To solve the quadratic equation, apply the quadratic formula: \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 4\), and \(c = -45\).
7. Calculate the discriminant: \({4}^2 - 4 \times 1 \times (-45) = 16 + 180 = 196\).
8. Since the discriminant is a perfect square, find the roots: \(y = \frac{-4 \pm \sqrt{196}}{2}\).
9. Compute \(\sqrt{196} = 14\).
10. Solve for \(y\):
    • First root: \(y = \frac{-4 + 14}{2} = \frac{10}{2} = 5\).
    • Second root: \(y = \frac{-4 - 14}{2} = \frac{-18}{2} = -9\).

Thus, the value of \(y\) is 5, which matches the given correct answer.