To solve the equation \( 4x^2 + 69 = 36x \) using the quadratic formula, we first rearrange it into standard form \( ax^2 + bx + c = 0 \).
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Rearranging the equation: \[ 4x^2 - 36x + 69 = 0 \]
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Identify the coefficients \( a \), \( b \), and \( c \):
- \( a = 4 \)
- \( b = -36 \)
- \( c = 69 \)
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Plug these coefficients into the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
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Calculate \( b^2 - 4ac \): \[ b^2 = (-36)^2 = 1296 \] \[ 4ac = 4 \cdot 4 \cdot 69 = 1104 \] \[ b^2 - 4ac = 1296 - 1104 = 192 \]
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Substitute back into the quadratic formula: \[ x = \frac{-(-36) \pm \sqrt{192}}{2 \cdot 4} \] \[ x = \frac{36 \pm \sqrt{192}}{8} \]
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Simplify \( \sqrt{192} \): \[ \sqrt{192} = \sqrt{64 \cdot 3} = 8\sqrt{3} \]
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Substitute \( \sqrt{192} \) back into the equation: \[ x = \frac{36 \pm 8\sqrt{3}}{8} \] \[ x = \frac{36}{8} \pm \frac{8\sqrt{3}}{8} \] \[ x = \frac{9}{2} \pm \sqrt{3} \]
Thus, the roots in simplest form are: \[ x = \frac{9}{2} + \sqrt{3} \quad \text{and} \quad x = \frac{9}{2} - \sqrt{3} \]