Question

The number of non-negative integer values of $k$ for which the quadratic equation $x^2 - 5x + k = 0$ has only integer roots, is:

Hint:

For quadratics with integer coefficients, to ensure integer roots: \begin{itemize} \item The discriminant must be a non-negative perfect square. \item Check that the resulting expression for the roots truly gives integers (often a parity check on the numerator). \end{itemize}

Answer 1

Correct Answer: 3

Approach: If the roots are integers, their sum and product are fixed by the equation: sum \(= 5\), product \(= k\). So just list integer pairs adding to 5, and each product is a valid \(k\). No discriminant juggling needed once you spot Vieta.

Step 1: Vieta's relations.
For \(x^2 - 5x + k = 0\) with integer roots \(p, q\):
\[ p + q = 5, \qquad pq = k. \]

Step 2: Need \(k \ge 0\) (non-negative).
So we want integer pairs \((p, q)\) with \(p + q = 5\) and \(pq \ge 0\).

Step 3: List the pairs and their products.
\((0,5) \to k = 0\); \((1,4) \to k = 4\); \((2,3) \to k = 6\).
Pairs like \((-1,6) \to k = -6\) or \((6,-1)\) give negative \(k\) — excluded. By symmetry \((5,0),(4,1),(3,2)\) repeat the same \(k\) values.

Step 4: Collect distinct non-negative \(k\).
\[ k \in \{0, 4, 6\}. \]
That is \(3\) values, each giving integer roots: \(\{0,5\}, \{1,4\}, \{2,3\}\).

Number of values = 3.

Answer 2

Approach: For \(x^2 - 5x + k = 0\) to have integer roots, its discriminant must be a perfect square. Turn this into a search over small perfect squares instead of listing root pairs directly.

Step 1: Discriminant condition. Roots are integers only if the discriminant \(D = 25 - 4k\) is a perfect square, say \(D = t^2\) for some integer \(t \ge 0\). Then \(k = \dfrac{25-t^2}{4}\).

Step 2: Find valid \(t\). We need \(k\) to be a non-negative integer, so \(25-t^2\) must be a non-negative multiple of 4. Checking \(t = 0,1,2,\ldots\): \(t=1 \Rightarrow k=\frac{24}{4}=6\); \(t=3 \Rightarrow k=\frac{16}{4}=4\); \(t=5 \Rightarrow k=\frac{0}{4}=0\). (Even \(t\) values like \(0,2,4\) give \(25-t^2\) not divisible by 4, and \(t\ge7\) makes \(k\) negative.)

Step 3: Count. The valid values are \(k=0,4,6\), three distinct non-negative integers.

\[ \boxed{3} \]