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11 Prove that
[tex] {x}^{2} + kx + 2 = 0[/tex]
has real roots if
[tex]k \geqslant 2 \sqrt{2} [/tex]
For which other values of k does the equation have real roots?​

Respuesta :

sqdancefan

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Answer:

  • the discriminant is non-negative for  k ≥ 2√2
  • k ≤ -2√2 will also give real roots

Step-by-step explanation:

The discriminant of this quadratic equation is ...

  d = b² -4ac

  d = k² -4(1)(2) = k² -8

For real roots, we require ...

  k² -8 ≥ 0

  k² ≥ 8

  |k| ≥ √8 = 2√2

That is, the equation will have real roots when ...

  k ≥ 2√2 . . . . . . . what you're asked to show

  k ≤ -2√2 . . . . . alternate values of k