Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

Determine the number and type of solution to the equation:?

Determine the number and type of solution to the equation:

8y^2 = 6y - 7

A) Exactly one real solution.

B) Exactly two real solutions.

C) Exactly two complex, but not real, solutions.

1 Answer

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  • ?
    Lv 7
    1 decade ago
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    You can find it by using determinant.

    First move everything to one side and set the other side equal to 0.

    -8y^2 + 6y - 7 =0

    Comparing this to the standard form, ay^2 + by + c =0, we see that a = -8, b = 6, and c = -7.

    Determinant D = b^2 - 4ac. Plug in the numbers to find D.

    D = b^2 - 4ac = 6^2 - 4(-8)(-7) = -188

    If D > 0, then you have two real different solutions.

    If D = 0, then you have one real solution (double root).

    If D < 0, then you have zero real solution, but two conjugate imaginary (complex) solutions.

    Since our D = -188 < 0, we have two imaginary (complex) solutions.

    So the correct answer is C).

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