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Which equations have no real solution but have two complex solutions?

Which equations have no real solution but have two complex solutions?-example-1

1 Answer

2 votes

Answer:

2x^2 = 6x - 5 and 3x^2 - 5x = -8

Explanation:

Well you can move around the terms in each equation to get it in the form ax^2 + bx + c. From there you can evaluate the value of the discriminant, which is the value under the radical in the quadratic equation. If the discriminant is negative that means you have two complex solutions. The discriminant is equal to b^2 -4ac

3x^2 - 5x = -8

3x^2 - 5x + 8 = 0

discriminant = (-5)^2 - 4(3)(8)

discriminant = 25 - 96

discriminant = -71

2x^2 = 6x - 5

0 = -2x^2 + 6x - 5

6^2 - 4(-2)(-5)

36 - 40

discriminant = -4

12x = 9x^2 + 4

0 = 9x^2 - 12x + 4

(-12)^2 - 4(9)(4)

144 - 144

0 (not negative, just one solution at the vertex)

-x^2 - 10x = 34

-x^2 - 10x + 34 = 0

(-10)^2 - 4(-1)(34)

100 + 136

236

User Vlad Bogdan
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