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Solve the following equation using the quadratic formula.

[tex]\[ 0 = -3n^2 + 27 \][/tex]

A. \{-2, 10\}
B. \left\{-3 \frac{1}{2}, 3 \frac{1}{2}\right\}
C. \left\{-2, 3 \frac{1}{2}\right\}
D. \{-3, 3\}

Sagot :

GinnyAnswer
Certainly! Let's solve the quadratic equation [tex]\( 0 = -3n^2 + 27 \)[/tex] using the quadratic formula step-by-step. The quadratic formula is:

[tex]\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]

The equation [tex]\( 0 = -3n^2 + 27 \)[/tex] can be rewritten in the standard quadratic form [tex]\( an^2 + bn + c = 0 \)[/tex], where:
- [tex]\( a = -3 \)[/tex]
- [tex]\( b = 0 \)[/tex]
- [tex]\( c = 27 \)[/tex]

Step-by-step solution:

1. Identify the coefficients [tex]\( a, b, \)[/tex] and [tex]\( c \)[/tex]:
[tex]\[ a = -3, \quad b = 0, \quad c = 27 \][/tex]

2. Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substituting the values of [tex]\( a, b, \)[/tex] and [tex]\( c \)[/tex]:
[tex]\[ \Delta = 0^2 - 4(-3)(27) \][/tex]
[tex]\[ \Delta = 0 + 324 \][/tex]
[tex]\[ \Delta = 324 \][/tex]

3. Since the discriminant is positive, we will have two real roots.

4. Calculate the roots using the quadratic formula:
[tex]\[ n = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
[tex]\[ n = \frac{-0 \pm \sqrt{324}}{2(-3)} \][/tex]
[tex]\[ n = \frac{\pm 18}{-6} \][/tex]

Simplifying the two possible solutions:
- For the positive square root:
[tex]\[ n_1 = \frac{18}{-6} = -3 \][/tex]
- For the negative square root:
[tex]\[ n_2 = \frac{-18}{-6} = 3 \][/tex]

Therefore, the roots of the equation [tex]\( 0 = -3n^2 + 27 \)[/tex] are:

[tex]\[ n = -3 \quad \text{and} \quad n = 3 \][/tex]

So, the correct answer is:

[tex]\[ \{-3, 3\} \][/tex]
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