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For what values of a does the equation ax^2+x+4=0 have only one real solution?

User Gwendal
by
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1 Answer

5 votes

Answer:

1/16

Explanation:

To have one real solution, the discriminant must be 0.

b² − 4ac = 0

1² − 4a(4) = 0

1 − 16a = 0

a = 1/16

User Carl Weis
by
6.4k points