Yefim S. answered • 07/09/20

Math Tutor with Experience

We have ax^{2} - b = - 2, ax^{2} = b - 2, x^{2} = (b - 2)/a. To have one solution (b - 2)/a = 0, so b - 2 = 0 and b = 2 and a ≠ 0. So right answer is (C): b = 2, a = - 2

Park H.

asked • 07/09/20y = -2

y = ax^2-b

(A). a=3, b=-3

(B). a=-2, b=-2

(C). a=-2, b=2

(D). a=3, b=3

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Yefim S. answered • 07/09/20

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(20)
Math Tutor with Experience

^{2} - b = - 2, ax^{2} = b - 2, x^{2} = (b - 2)/a. To have one solution (b - 2)/a = 0, so b - 2 = 0 and b = 2 and a ≠ 0. So right answer is (C): b = 2, a = - 2

Billy W. answered • 07/09/20

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The problem states that this is a system of equations so we can write it as 1 equation:

-2=ax^{2}-b

Now let's plug in the given values

a) -2=3x^2+3

-5=3x^2

-5/3=x^2

sqrt(-5/3)=x

0 (real) solutions

B) -2=(-2)x^2+2

-4=-2x^2

2=x^2

x=+/- Sqrt(2)

2 real solutions

C)-2=-2x^2-2

0=-2x^2

0=x^2

x=0

1 Real solution

C is the answer

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