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NDA 2026
Exam
NDA
Year
2026
Subject
Mathematics
Chapter
Quadratic Equation

NDA 2026 · Mathematics · Quadratic Equation

For a given k , what is the minimum value of x^2+kx+k^2 ?

For a given kk, what is the minimum value of x2+kx+k2x^2+kx+k^2 ?

Options

  1. A00
  2. Bk24\dfrac{k^2}{4}
  3. C3k24\dfrac{3k^2}{4}
  4. Dk22\dfrac{k^2}{2}

Correct answer

C. 3k24\dfrac{3k^2}{4}

Step-by-step solution

Let the given expression be f(x)=x2+kx+k2f(x) = x^2 + kx + k^2.

To find the minimum value, we can complete the square for the quadratic expression in xx:

f(x)=x2+kx+k24−k24+k2f(x) = x^2 + kx + \dfrac{k^2}{4} - \dfrac{k^2}{4} + k^2

f(x)=(x+k2)2+3k24f(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 0forallrealfor all realx.<br><br>Theminimumvalueof.<br><br>The minimum value off(x)occurswhenoccurs whenx = -\dfrac{k}{2}.<br><br>Therefore,theminimumvalueoftheexpressionis.<br><br>Therefore, the minimum value of the expression is\dfrac{3k^2}{4}.<br><br>Answer:.<br><br>Answer:\dfrac{3k^2}{4}
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