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Find a value of b such that 2x^2 + bx + 8 will have only 1 real solution.

a. b = -4
b. b = 4
c. b = 8
d. b = -8

User Besen
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1 Answer

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Final answer:

The value of b that makes the quadratic equation 2x^2 + bx + 8 have one real solution is either b = 8 or b = -8, which are the values that make the discriminant equal zero.

Step-by-step explanation:

To find a value of b such that the quadratic equation 2x^2 + bx + 8 has only one real solution, we need to ensure that the discriminant of the equation, given by b^2 - 4ac, equals zero. The equation takes the general form ax^2 + bx + c = 0, where a = 2, b is the unknown we are solving for, and c = 8.

Applying the discriminant formula, we want b^2 - 4ac = 0, which simplifies to b^2 - 4(2)(8) = 0 or b^2 - 64 = 0. Solving for b, we find b = ±8. Looking at the options provided, the value for b that results in one real solution, meeting these conditions, is b = 8 or b = -8.

User Mahmmoud Kinawy
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