MCQ
The equation $x^2-8 x+k=0$ has real and distinct roots if :
  • A
    $k = 16$
  • B
    $k > 16$
  • C
    $k = 8$
  • $k < 16$

Answer

Correct option: D.
$k < 16$
Since the given equation has real roots,
i.e., $D=b^2-4 a c=0$
Here $, a = 1, b = -8, c = k$
$\therefore(-8)^2-4(1)(\text{k})\geq0$
Or, $64-4\text{k}\geq0$
$4\text{k}\leq64$
OR
$\text{k}\leq\frac{64}{4}$
Or, $\text{k}\leq16$

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