QuestionSolve the equation by completing the square. $3 n^{2}=-6 n-1$

Studdy Solution

STEP 1

1. The equation $3n^2 = -6n - 1$ is a quadratic equation.
2. We will solve the equation by completing the square, which involves manipulating the equation into a perfect square trinomial.

STEP 2

1. Move all terms to one side of the equation.
2. Divide by the leading coefficient to simplify.
3. Complete the square.
4. Solve for $n$ .

STEP 3

Move all terms to one side of the equation to set it equal to zero:
$3n^2 + 6n + 1 = 0$

STEP 4

Divide the entire equation by the leading coefficient (3) to simplify the quadratic term:
$n^2 + 2n + \frac{1}{3} = 0$

STEP 5

To complete the square, focus on the $n^2 + 2n$ part. Take half of the coefficient of $n$ , square it, and add and subtract it inside the equation:
The coefficient of $n$ is 2. Half of 2 is 1, and squaring it gives $1^2 = 1$ .
$n^2 + 2n + 1 - 1 + \frac{1}{3} = 0$

STEP 6

Rewrite the equation as a perfect square trinomial and simplify the constant terms:
$(n + 1)^2 - 1 + \frac{1}{3} = 0$
Combine the constants:
$(n + 1)^2 - \frac{2}{3} = 0$

STEP 7

Solve for $n$ by isolating the perfect square and taking the square root of both sides:
$(n + 1)^2 = \frac{2}{3}$
Take the square root of both sides:
$n + 1 = \pm \sqrt{\frac{2}{3}}$

STEP 8

Solve for $n$ by subtracting 1 from both sides:
$n = -1 \pm \sqrt{\frac{2}{3}}$
The solutions for $n$ are:
$n = -1 + \sqrt{\frac{2}{3}} \quad \text{and} \quad n = -1 - \sqrt{\frac{2}{3}}$

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