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In the given equation bx²+3x+2=0, where b is a constant. For which of the following values of b will the equation have more than one real solution?

a) b=1
b) b=−2
c) b=0
d) b=3

1 Answer

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Final answer:

The quadratic equation bx²+3x+2=0 will have more than one real solution if the discriminant (b² - 4ac) is positive. After evaluating the discriminant for each value of b, the equation has more than one real solution for b=1 and b=-2.

Step-by-step explanation:

To determine for which values of b the quadratic equation bx²+3x+2=0 will have more than one real solution, we must look at the equation's discriminant, which is found using the formula b² - 4ac. In this context, a is the coefficient of x² (which is b), b is the coefficient of x (which is 3), and c is the constant term (which is 2).

A quadratic equation will have more than one real solution if the discriminant is positive, which means b² - 4ac > 0. Let's analyze the discriminant for each given value of b:

  • For b=1: (3² - 4(1)(2)) = 9 - 8 = 1. Since 1 is positive, there will be more than one real solution.
  • For b=-2: (3² - 4(-2)(2)) = 9 - (-16) = 25. Since 25 is positive, there will be more than one real solution.
  • For b=0: The equation becomes linear (3x + 2 = 0), so it only has one solution.
  • For b=3: (3² - 4(3)(2)) = 9 - 24 = -15. Since -15 is negative, there will be no real solutions.

Therefore, the quadratic equation will have more than one real solution for b=1 and b=-2.

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