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505-555 (Easy)|   Algebra|   Exponents|      
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Bunuel
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Suppose that x = -2 is the only number that satisfices 2^x^2 + Bx 07 Aug 2023, 06:00
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C
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15% (low)

Question Stats:

84% (01:48) correct 16% (02:48) wrong based on 1030 sessions

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Suppose that x = -2 is the only number that satisfies \(2^{x^2 + Bx + \frac{11}{2}} = \sqrt{8}\). What is the value of B?

A. 3/4
B. 15/4
C. 4
D. 9/2
E. 19/4­
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C
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Suppose that x = -2 is the only number that satisfices 2^x^2 + Bx 07 Aug 2023, 07:05
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Bunuel
Suppose x = -2 is the only number that satisfices \(2^{x^2 + Bx + \frac{11}{2}} = \sqrt{8}\). What is the value of B?

A. 3/4
B. 15/4
C. 4
D. 9/2
E. 19/4
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We can equate powers on both sides of the equation and obtain the value of B.

Option C
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Suppose that x = -2 is the only number that satisfices 2^x^2 + Bx 07 Aug 2023, 07:20
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Bunuel
Suppose x = -2 is the only number that satisfices \(2^{x^2 + Bx + \frac{11}{2}} = \sqrt{8}\). What is the value of B?

A. 3/4
B. 15/4
C. 4
D. 9/2
E. 19/4
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√8 = (2^3)^(1/2) = 2^(3/2)

We see that each side of the equation in the question stem can be written as 2 raised to some exponent.

If two exponential expressions that have the same base are equal and the same base is not -1, 0, or 1, then their exponents must be equal as well.

x^2 + Bx + 11/2 = 3/2

Since x = -2, we have:

(-2)^2 – 2B + 11/2 = 3/2

8 – 4B + 11 = 3

16 = 4B

B = 4

Answer: C
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Suppose that x = -2 is the only number that satisfices 2^x^2 + Bx 17 Mar 2025, 20:15
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Classic exponents situation. Make the bases equal and then set the exponents equal:
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Suppose that x = -2 is the only number that satisfices 2^x^2 + Bx 18 Mar 2025, 04:30
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Bunuel
Suppose that x = -2 is the only number that satisfies \(2^{x^2 + Bx + \frac{11}{2}} = \sqrt{8}\). What is the value of B?

A. 3/4
B. 15/4
C. 4
D. 9/2
E. 19/4­
Show more
TRICK: It's important to bring both sides in same base so that their exponents may be equated. So in this question it's best to bring the right side in form of \(2^a\)

We know, \(2^{x^2 + Bx + \frac{11}{2}} = \sqrt{8}\)

but we also know that \(\sqrt{8} = 2^{3/2}\)

therefore, \(2^{x^2 + Bx + \frac{11}{2}} = 2^{3/2}\)

i.e. \(x^2 + Bx + \frac{11}{2} = \frac{3}{2}\)

i.e. \(x^2 + Bx + 4 = 0\)

Now, we have been given that x = -2

i.e. \((-2)^2 + B*(-2) + 4 = 0\)

i.e. B = 4

Answer: Option C

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Suppose that x = -2 is the only number that satisfices 2^x^2 + Bx 08 Oct 2025, 11:37
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Cant i write root 8as 2 ^1 * 2^ 1/2
and hence
8 - 4B + 11 = 1?
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