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Question

Find the value of $$k$$ for which the equation $$3{x}^{2}-6x+k=0$$ has distinct and real root.


Solution

$$3{x^2} - 6x + k = 0$$
$$For\,real\,distinct\,roots$$
$$D \ge 0$$
$${b^2} - 4ac \ge 0$$
$${\left( { - 6} \right)^2} - 4 \times 3 \times k \ge 0$$
$$36 - 12k \ge 0$$
$$36 \ge 12k$$
$$k \le 3$$

Mathematics

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