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What are two possible integers to make the equation ax² + 10x + c = 0 have two real solutions?

User Woodbase
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Final answer:

Two possible integer pairs for the quadratic equation ax² + 10x + c = 0 to have two real solutions, based on the discriminant being non-negative, could be (1, 25) or (2, 12).

Step-by-step explanation:

To determine two possible integers for a and c to make the quadratic equation ax² + 10x + c = 0 have two real solutions, we should consider the discriminant, which is the part of the quadratic formula under the radical, represented by b² - 4ac.

For real solutions to exist, the discriminant must be greater than or equal to zero.

We have the value of b as 10 in our equation.

Using the discriminant condition, we can set up the inequality 10² - 4ac ≥ 0, which simplifies to 100 - 4ac ≥ 0.

From there, we can solve for c in terms of a, or vice versa, to find suitable values.

If a is 1, then c must be less than or equal to 25.

If a is 2, then c must be less than or equal to 12.5 and so on.

Therefore, two such integer pairs could be (a, c) = (1, 25) or (2, 12).

User Endran
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