x 2 - 12 x + 27 = 0 How many distinct real solutions does the given equation have?

Question

Accepted Answer

Choice A is correct. The number of solutions of a quadratic equation of the form $a x^{2} + b x + c = 0$ , where $a$ , $b$ , and $c$ are constants, can be determined by the value of the discriminant, $b squared minus 4 a c$ . If the value of the discriminant is positive, then the quadratic equation has exactly two distinct real solutions. If the value of the discriminant is equal to zero, then the quadratic equation has exactly one real solution. If the value of the discriminant is negative, then the quadratic equation has zero real solutions. In the given equation, $x^{2} - 12 x + 27 = 0$ , $a equals 1$ , $b = - 12$ , and $c equals 27$ . Substituting these values for $a$ , $b$ , and $c$ in $b squared minus 4 a c$ yields ${(- 12)}^{2} - 4 (1) (27)$ , or $36$ . Since the value of its discriminant is positive, the given equation has exactly two distinct real solutions.

Choice B is incorrect and may result from conceptual or calculation errors.

Choice C is incorrect and may result from conceptual or calculation errors.

Choice D is incorrect and may result from conceptual or calculation errors.

Question x-2-12

x 2 - 12 x + 27 = 0 How many distinct real solutions does the

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