If a is an integer, for how many integral values of n can the quadratic equation x2 – (2a + 3)x + 4n = 0, have real and equal roots for x?
Question
Solution 1
The quadratic equation is given as x^2 - (2a + 3)x + 4n = 0.
For a quadratic equation ax^2 + bx + c = 0 to have real and equal roots, the discriminant (b^2 - 4ac) must be equal to zero.
In this case, a = 1, b = -(2a + 3), and c = 4n.
So, we have [-(2a + 3)]^2 - 414n = 0.
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