What is b if y= x^2 + 3x + b has two complex roots.?

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A quadratic equation ax^2 + bx + c = 0 has two complex roots if b^2 - 4ac < 0

In the given equation x^2 + 3x + b,

a = 1 , b = 3 and c = b( to be determined)

Now as b^2 - 4ac = 9 - 4b < 0

=> 4b > 9

=> b > 9/4

Therefore b > 9/4

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Given the quadratic equation :

y= x^2 + 3x + b

We need to determine b such that y has two complex roots.

We know that if delta < 0, then the quadratic equation has 2 complex roots.

==> delta = b^2 - 4ac < 0

==> a = 1   b= 3    c = b

==> 9 - 4*1*b < 0

==> 9- 4b < 0

==> -4b < -9

==> b > 9/4

Then the values of b should be greater that 9/4

Then, b belongs to the interval ( 9/4 , inf )

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