Solve by Completing the Square 4x^2-12x+9=0

$4 x^{2} - 12 x + 9 = 0$

Subtract $9$ from both sides of the equation.

$4 x^{2} - 12 x = - 9$

Divide each term in $4 x^{2} - 12 x = - 9$ by $4$ and simplify.

Divide each term in $4 x^{2} - 12 x = - 9$ by $4$ .

$\frac{4 x^{2}}{4} + \frac{- 12 x}{4} = \frac{- 9}{4}$

Simplify the left side.

Simplify each term.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x^{2}}{4} + \frac{- 12 x}{4} = \frac{- 9}{4}$

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

$x^{2} + \frac{- 12 x}{4} = \frac{- 9}{4}$

Cancel the common factor of $- 12$ and $4$ .

Factor $4$ out of $- 12 x$ .

$x^{2} + \frac{4 (- 3 x)}{4} = \frac{- 9}{4}$

Cancel the common factors.

Factor $4$ out of $4$ .

$x^{2} + \frac{4 (- 3 x)}{4 (1)} = \frac{- 9}{4}$

Cancel the common factor.

$x^{2} + \frac{4 (- 3 x)}{4 \cdot 1} = \frac{- 9}{4}$

Rewrite the expression.

$x^{2} + \frac{- 3 x}{1} = \frac{- 9}{4}$

Divide $- 3 x$ by $1$ .

$x^{2} - 3 x = \frac{- 9}{4}$

Simplify the right side.

Move the negative in front of the fraction.

$x^{2} - 3 x = - \frac{9}{4}$

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

${(\frac{b}{2})}^{2} = {(- \frac{3}{2})}^{2}$

Add the term to each side of the equation.

$x^{2} - 3 x + {(- \frac{3}{2})}^{2} = - \frac{9}{4} + {(- \frac{3}{2})}^{2}$

Simplify the equation.

Simplify the left side.

Simplify each term.

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $- \frac{3}{2}$ .

$x^{2} - 3 x + {(- 1)}^{2} {(\frac{3}{2})}^{2} = - \frac{9}{4} + {(- \frac{3}{2})}^{2}$

Apply the product rule to $\frac{3}{2}$ .

$x^{2} - 3 x + {(- 1)}^{2} \frac{3^{2}}{2^{2}} = - \frac{9}{4} + {(- \frac{3}{2})}^{2}$

Raise $- 1$ to the power of $2$ .

$x^{2} - 3 x + 1 \frac{3^{2}}{2^{2}} = - \frac{9}{4} + {(- \frac{3}{2})}^{2}$

Multiply $\frac{3^{2}}{2^{2}}$ by $1$ .

$x^{2} - 3 x + \frac{3^{2}}{2^{2}} = - \frac{9}{4} + {(- \frac{3}{2})}^{2}$

Raise $3$ to the power of $2$ .

$x^{2} - 3 x + \frac{9}{2^{2}} = - \frac{9}{4} + {(- \frac{3}{2})}^{2}$

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

$x^{2} - 3 x + \frac{9}{4} = - \frac{9}{4} + {(- \frac{3}{2})}^{2}$

Simplify the right side.

Simplify $- \frac{9}{4} + {(- \frac{3}{2})}^{2}$ .

Simplify each term.

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $- \frac{3}{2}$ .

$x^{2} - 3 x + \frac{9}{4} = - \frac{9}{4} + {(- 1)}^{2} {(\frac{3}{2})}^{2}$

Apply the product rule to $\frac{3}{2}$ .

$x^{2} - 3 x + \frac{9}{4} = - \frac{9}{4} + {(- 1)}^{2} \frac{3^{2}}{2^{2}}$

Raise $- 1$ to the power of $2$ .

$x^{2} - 3 x + \frac{9}{4} = - \frac{9}{4} + 1 \frac{3^{2}}{2^{2}}$

Multiply $\frac{3^{2}}{2^{2}}$ by $1$ .

$x^{2} - 3 x + \frac{9}{4} = - \frac{9}{4} + \frac{3^{2}}{2^{2}}$

Raise $3$ to the power of $2$ .

$x^{2} - 3 x + \frac{9}{4} = - \frac{9}{4} + \frac{9}{2^{2}}$

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

$x^{2} - 3 x + \frac{9}{4} = - \frac{9}{4} + \frac{9}{4}$

Combine fractions.

Combine the numerators over the common denominator.

$x^{2} - 3 x + \frac{9}{4} = \frac{- 9 + 9}{4}$

Simplify the expression.

Add $- 9$ and $9$ .

$x^{2} - 3 x + \frac{9}{4} = \frac{0}{4}$

Divide $0$ by $4$ .

$x^{2} - 3 x + \frac{9}{4} = 0$

Factor the perfect trinomial square into ${(x - \frac{3}{2})}^{2}$ .

${(x - \frac{3}{2})}^{2} = 0$

Solve the equation for $x$ .

Set the $x - \frac{3}{2}$ equal to $0$ .

$x - \frac{3}{2} = 0$

Add $\frac{3}{2}$ to both sides of the equation.

$x = \frac{3}{2}$

Solve by Completing the Square 4x^2-12x+9=0

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