Q192 / 270 · NDA 2026 (Phase 1) · Mathematics
For a given $k$, what is the minimum value of $x^2+kx+k^2$ ?
- A$0$
- B$\dfrac{k^2}{4}$
- Correct answerC$\dfrac{3k^2}{4}$
- D$\dfrac{k^2}{2}$
Solution
Let the given expression be $f(x) = x^2 + kx + k^2$.
To find the minimum value, we can complete the square for the quadratic expression in $x$:
$f(x) = x^2 + kx + \dfrac{k^2}{4} - \dfrac{k^2}{4} + k^2$
$f(x) = \left(x + \dfrac{k}{2}\right)^2 + \dfrac{3k^2}{4}$
Since the square of a real number is always non-negative, $\left(x + \dfrac{k}{2}\right)^2 \ge 0$ for all real $x$.
The minimum value of $f(x)$ occurs when $x = -\dfrac{k}{2}$.
Therefore, the minimum value of the expression is $\dfrac{3k^2}{4}$.
Answer: $\dfrac{3k^2}{4}$
To find the minimum value, we can complete the square for the quadratic expression in $x$:
$f(x) = x^2 + kx + \dfrac{k^2}{4} - \dfrac{k^2}{4} + k^2$
$f(x) = \left(x + \dfrac{k}{2}\right)^2 + \dfrac{3k^2}{4}$
Since the square of a real number is always non-negative, $\left(x + \dfrac{k}{2}\right)^2 \ge 0$ for all real $x$.
The minimum value of $f(x)$ occurs when $x = -\dfrac{k}{2}$.
Therefore, the minimum value of the expression is $\dfrac{3k^2}{4}$.
Answer: $\dfrac{3k^2}{4}$
Hence, the correct answer is option: 'C'.