Multiple choice

For what real values of $k$, other than $k=0$, does the equation ${ x }^{ 2 }+kx+{ k }^{ 2 }=0$ have real roots? (The symbol $x\ge a$ means that $x$ can take on all values greater than $a$ and the value $a$ itself; $x\le a$ has the corresponding meaning with "less than")

  1. $k< 0$
  2. $k> 0$
  3. $k\ge 1$
  4. all values of $k$
  5. no values of $k$
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E Correct answer
Explanation

For the quadratic equation x^2 + kx + k^2 = 0 to have real roots, the discriminant D = b^2 - 4ac must be greater than or equal to 0. Here, D = k^2 - 4(1)(k^2) = -3k^2. Since k is not 0, -3k^2 is always negative, meaning there are no real roots for any non-zero k.

AI explanation

For the equation x^2 + kx + k^2 = 0 to have real roots, the discriminant (b^2 - 4ac) must be greater than or equal to zero. Substituting a = 1, b = k, and c = k^2 gives k^2 - 4k^2, which equals -3k^2. Since the square of any real number is non-negative, -3k^2 is always less than or equal to zero, and is strictly zero only when k = 0; therefore, there are no values of k other than zero that produce real roots.

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