Comp_Geek_573 wrote:

(True/False)

In the following theorem, x is a free variable:

Suppose a, b and c are positive real numbers and b^2 > 4ac. Then the equation ax^2 + bx + c = 0 always has two solutions.

Let a=c=1, b=2

ax^2 + bx +c = (x+1)^2, which has only one solution, x=-1

So it would have been false anyway. Personally I think the question was worded poorly.

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