The number of real values of k for which the equation 2x²-6x+ k = 0 has two distinct roots in [0, 1] is041infinitely many
Question
Solution 1
The given quadratic equation is 2x² - 6x + k = 0.
For a quadratic equation ax² + bx + c = 0 to have two distinct real roots, the discriminant (b² - 4ac) must be greater than 0.
In this case, a = 2, b = -6, and c = k. So, the discriminant is (-6)² - 42k = 36 - 8k.
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