Solve by Completing the Square 4x^2-16x+12=0

$4 x^{2} - 16 x + 12 = 0$

Step 1

Subtract $12$ from both sides of the equation.

$4 x^{2} - 16 x = - 12$

Step 2

Divide each term in $4 x^{2} - 16 x = - 12$ by $4$ and simplify.

Divide each term in $4 x^{2} - 16 x = - 12$ by $4$ .

$\frac{4 x^{2}}{4} + \frac{- 16 x}{4} = \frac{- 12}{4}$

Simplify the left side.

Simplify each term.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x^{2}}{4} + \frac{- 16 x}{4} = \frac{- 12}{4}$

Divide $x^{2}$ by $1$ .

$x^{2} + \frac{- 16 x}{4} = \frac{- 12}{4}$

Cancel the common factor of $- 16$ and $4$ .

Factor $4$ out of $- 16 x$ .

$x^{2} + \frac{4 (- 4 x)}{4} = \frac{- 12}{4}$

Cancel the common factors.

Factor $4$ out of $4$ .

$x^{2} + \frac{4 (- 4 x)}{4 (1)} = \frac{- 12}{4}$

Cancel the common factor.

$x^{2} + \frac{4 (- 4 x)}{4 \cdot 1} = \frac{- 12}{4}$

Rewrite the expression.

$x^{2} + \frac{- 4 x}{1} = \frac{- 12}{4}$

Divide $- 4 x$ by $1$ .

$x^{2} - 4 x = \frac{- 12}{4}$

Simplify the right side.

Divide $- 12$ by $4$ .

$x^{2} - 4 x = - 3$

Step 3

To create a trinomial square on the left side of the equation, find a value that is equal to the square of half of $b$ .

${(\frac{b}{2})}^{2} = {(- 2)}^{2}$

Step 4

Add the term to each side of the equation.

$x^{2} - 4 x + {(- 2)}^{2} = - 3 + {(- 2)}^{2}$

Step 5

Simplify the equation.

Simplify the left side.

Raise $- 2$ to the power of $2$ .

$x^{2} - 4 x + 4 = - 3 + {(- 2)}^{2}$

Simplify the right side.

Simplify $- 3 + {(- 2)}^{2}$ .

Raise $- 2$ to the power of $2$ .

$x^{2} - 4 x + 4 = - 3 + 4$

Add $- 3$ and $4$ .

$x^{2} - 4 x + 4 = 1$

Step 6

Factor the perfect trinomial square into ${(x - 2)}^{2}$ .

${(x - 2)}^{2} = 1$

Step 7

Solve the equation for $x$ .

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x - 2 = \pm \sqrt{1}$

Any root of $1$ is $1$ .

$x - 2 = \pm 1$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x - 2 = 1$

Move all terms not containing $x$ to the right side of the equation.

Add $2$ to both sides of the equation.

$x = 1 + 2$

Add $1$ and $2$ .

$x = 3$

Next, use the negative value of the $\pm$ to find the second solution.

$x - 2 = - 1$

Move all terms not containing $x$ to the right side of the equation.

Add $2$ to both sides of the equation.

$x = - 1 + 2$

Add $- 1$ and $2$ .

$x = 1$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 3, 1$

Solve by Completing the Square 4x^2-16x+12=0

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