Solve for x 2x^2-12x=-6

The problem provided is a quadratic equation that requires you to find the value(s) of the variable x that satisfy the equation. Specifically, you are asked to isolate x and determine which number(s) can be plugged in for x to make the equation true. To solve this equation, you can either factorize the quadratic expression, complete the square, or use the quadratic formula, after first reorganizing the equation to the standard form ax^2 + bx + c = 0 by adding 6 to both sides of the equation.

$2 x^{2} - 12 x = - 6$

Answer

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Solution:

Step 1:

Move $6$ to the right side to set the quadratic equation to zero: $2 x^{2} - 12 x + 6 = 0$ .

Step 2:

Extract the common factor of $2$ from the quadratic expression.

Step 2.1:

Take $2$ out of $2 x^{2}$ : $2 (x^{2}) - 12 x + 6 = 0$ .

Step 2.2:

Take $2$ out of $- 12 x$ : $2 (x^{2}) + 2 (- 6 x) + 6 = 0$ .

Step 2.3:

Take $2$ out of $6$ : $2 (x^{2}) + 2 (- 6 x) + 2 \cdot 3 = 0$ .

Step 2.4:

Combine the terms with a common factor of $2$ : $2 (x^{2} - 6 x) + 2 \cdot 3 = 0$ .

Step 2.5:

Factor out the $2$ completely: $2 (x^{2} - 6 x + 3) = 0$ .

Step 3:

Divide the equation by $2$ to simplify.

Step 3.1:

Divide each term by $2$ : $\frac{2 (x^{2} - 6 x + 3)}{2} = \frac{0}{2}$ .

Step 3.2:

Simplify the left side by canceling out the $2$ .

Step 3.2.1:

Cancel out the $2$ : $\frac{2 (x^{2} - 6 x + 3)}{2} = \frac{0}{2}$ .

Step 3.2.1.1:

The equation simplifies to $x^{2} - 6 x + 3 = 0$ .

Step 3.3:

Simplify the right side: $x^{2} - 6 x + 3 = 0$ .

Step 4:

Apply the quadratic formula: $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ .

Step 5:

Insert the coefficients $a = 1$ , $b = - 6$ , and $c = 3$ into the formula.

Step 6:

Carry out the arithmetic operations.

Step 6.1:

Work on the numerator.

Step 6.1.1:

Square $- 6$ : $x = \frac{6 \pm \sqrt{36 - 4 \cdot 1 \cdot 3}}{2 \cdot 1}$ .

Step 6.1.2:

Calculate $4 \cdot 1 \cdot 3$ .

Step 6.1.2.1:

Multiply $- 4$ by $1$ .

Step 6.1.2.2:

Multiply $- 4$ by $3$ : $x = \frac{6 \pm \sqrt{36 - 12}}{2 \cdot 1}$ .

Step 6.1.3:

Subtract $12$ from $36$ : $x = \frac{6 \pm \sqrt{24}}{2 \cdot 1}$ .

Step 6.1.4:

Express $24$ as $2^{2} \cdot 6$ .

Step 6.1.4.1:

Factor out $4$ from $24$ : $x = \frac{6 \pm \sqrt{4 (6)}}{2 \cdot 1}$ .

Step 6.1.4.2:

Rewrite $4$ as $2^{2}$ : $x = \frac{6 \pm \sqrt{2^{2} \cdot 6}}{2 \cdot 1}$ .

Step 6.1.5:

Extract terms from under the radical: $x = \frac{6 \pm 2 \sqrt{6}}{2 \cdot 1}$ .

Step 6.2:

Multiply $2$ by $1$ : $x = \frac{6 \pm 2 \sqrt{6}}{2}$ .

Step 6.3:

Simplify the fraction: $x = 3 \pm \sqrt{6}$ .

Step 7:

Combine the solutions: $x = 3 + \sqrt{6}, 3 - \sqrt{6}$ .

Step 8:

Present the result in different forms.

Exact Form: $x = 3 + \sqrt{6}, 3 - \sqrt{6}$ .

Decimal Form: $x \approx 5.44948974, x \approx 0.55051025$ .

Knowledge Notes:

To solve the quadratic equation $2 x^{2} - 12 x = - 6$ , we follow a systematic approach:

Setting the Equation to Zero: The first step in solving a quadratic equation is to set it to zero by moving all terms to one side.
Factoring Out Common Terms: If there is a common factor in all terms of the quadratic equation, it should be factored out to simplify the equation.
Simplifying the Equation: After factoring, the equation can often be simplified by dividing both sides by the common factor.
Quadratic Formula: The quadratic formula $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ is a standard method for solving quadratic equations, where $a$ , $b$ , and $c$ are the coefficients of the terms $a x^{2}$ , $b x$ , and $c$ , respectively.
Arithmetic Operations: The process involves arithmetic operations such as squaring numbers, multiplying, adding, subtracting, and taking square roots.
Simplification of Radical Expressions: When dealing with square roots, it is often useful to factor the number under the radical to simplify the expression.
Final Solutions: The quadratic formula can yield two solutions, which are the roots of the equation.
Multiple Forms of the Result: Solutions can be presented in exact form (with radicals) or in decimal form (approximations).

Understanding these steps is crucial for solving quadratic equations effectively.