Question
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Rewrite the equation by completing the square. \( 4 x^{2}-12 x+9=0 \) \( (x+\square)^{2}=\square \)

Ask by Hardy Murphy. in the United States
Jan 24,2025

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Tutor-Verified Answer

Answer

\[ \left(x - \frac{3}{2}\right)^2 = 0 \]

Solution

To complete the square for the equation \( 4x^{2} - 12x + 9 = 0 \), follow these steps: 1. **Factor out the coefficient of \( x^2 \):** \[ 4(x^{2} - 3x) + 9 = 0 \] 2. **Complete the square inside the parentheses:** - Take half of the coefficient of \( x \), which is \( -\frac{3}{2} \). - Square it to get \( \left(-\frac{3}{2}\right)^2 = \frac{9}{4} \). Add and subtract this value inside the parentheses: \[ 4\left(x^{2} - 3x + \frac{9}{4} - \frac{9}{4}\right) + 9 = 0 \] \[ 4\left(\left(x - \frac{3}{2}\right)^2 - \frac{9}{4}\right) + 9 = 0 \] 3. **Simplify the equation:** \[ 4\left(x - \frac{3}{2}\right)^2 - 9 + 9 = 0 \] \[ 4\left(x - \frac{3}{2}\right)^2 = 0 \] 4. **Divide both sides by 4 to isolate the squared term:** \[ \left(x - \frac{3}{2}\right)^2 = 0 \] So, the equation rewritten by completing the square is: \[ \left(x - \frac{3}{2}\right)^{2} = 0 \] **Final Answer:** \[ \left(x - \dfrac{3}{2}\right)^{2} = 0 \]

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To rewrite the equation \( 4x^{2} - 12x + 9 = 0 \) by completing the square, we start by factoring out 4 from the first two terms: \[ 4(x^{2} - 3x) + 9 = 0. \] Next, to complete the square inside the parentheses, we take half of the coefficient of \( x \) which is -3, square it to get \( \left(-\frac{3}{2}\right)^{2} = \frac{9}{4} \). We add and subtract this value within the parentheses: \[ 4 \left( x^{2} - 3x + \frac{9}{4} - \frac{9}{4} \right) + 9 = 0. \] This simplifies to: \[ 4 \left( \left( x - \frac{3}{2} \right)^{2} - \frac{9}{4} \right) + 9 = 0. \] Distributing the 4 gives: \[ 4 \left( x - \frac{3}{2} \right)^{2} - 9 + 9 = 0, \] which simplifies to: \[ 4 \left( x - \frac{3}{2} \right)^{2} = 0. \] So, rewriting it gives us: \[ \left( x - \frac{3}{2} \right)^{2} = 0. \] Thus completing the square results in: \[ (x - \frac{3}{2})^{2} = 0. \] So, \( \square = -\frac{3}{2} \) and \( \square = 0 \).

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