3x^2 + 36x + 81 = 0

Q: 3x^2 + 36x + 81 = 0

To solve the quadratic equation $3x^2 + 36x + 81 = 0$, we can use the quadratic formula, which is given by: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] where $a$, $b$, and $c$ are the coefficients from the quadratic equation $ax^2 + bx + c = 0$. In this case, $a = 3$, $b = 36$, and $c = 81$. Let's plug these values into the quadratic formula: \[x = \frac{-36 \pm \sqrt{36^2 - 4 \cdot 3 \cdot 81}}{2 \cdot 3}\] \[x = \frac{-36 \pm \sqrt{1296 - 972}}{6}\] \[x = \frac{-36 \pm \sqrt{324}}{6}\] \[x = \frac{-36 \pm 18}{6}\] There are two possible solutions based on the $\pm$ sign: 1. For $+$: \[x = \frac{-36 + 18}{6} = \frac{-18}{6} = -3\] 2. For $-$: \[x = \frac{-36 - 18}{6} = \frac{-54}{6} = -9\] Therefore, the solutions to the equation $3x^2 + 36x + 81 = 0$ are $x = -3$ and $x = -9$.

Question

3x^2 + 36x + 81  = 0

Accepted Answer

To solve the quadratic equation $3x^2 + 36x + 81 = 0$, we can use the quadratic formula, which is given by:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

where $a$, $b$, and $c$ are the coefficients from the quadratic equation $ax^2 + bx + c = 0$. In this case, $a = 3$, $b = 36$, and $c = 81$.

Let's plug these values into the quadratic formula:

$$x = \frac{-36 \pm \sqrt{36^2 - 4 \cdot 3 \cdot 81}}{2 \cdot 3}$$

$$x = \frac{-36 \pm \sqrt{1296 - 972}}{6}$$

$$x = \frac{-36 \pm \sqrt{324}}{6}$$

$$x = \frac{-36 \pm 18}{6}$$

There are two possible solutions based on the $\pm$ sign:

1. For $+$:

$$x = \frac{-36 + 18}{6} = \frac{-18}{6} = -3$$

2. For $-$:

$$x = \frac{-36 - 18}{6} = \frac{-54}{6} = -9$$

Therefore, the solutions to the equation $3x^2 + 36x + 81 = 0$ are $x = -3$ and $x = -9$.