a) \( 2 x(4-3 x)=(3+2 x)(3-2 x) \)
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To solve the equation \( 2x(4 - 3x) = (3 + 2x)(3 - 2x) \), let's start by expanding both sides. On the left side: \[ 2x(4 - 3x) = 8x - 6x^2. \] On the right side, we can use the difference of squares formula: \[ (3 + 2x)(3 - 2x) = 3^2 - (2x)^2 = 9 - 4x^2. \] Now we equate the two sides: \[ 8x - 6x^2 = 9 - 4x^2. \] Next, we can rearrange the equation to bring all terms to one side: \[ -6x^2 + 4x^2 + 8x - 9 = 0 \] which simplifies to: \[ -2x^2 + 8x - 9 = 0. \] To make calculations easier, let's multiply the entire equation by -1: \[ 2x^2 - 8x + 9 = 0. \] Now, we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Setting \( a = 2, b = -8, c = 9 \): 1. Calculate the discriminant: \[ b^2 - 4ac = (-8)^2 - 4 \cdot 2 \cdot 9 = 64 - 72 = -8. \] Since the discriminant is negative, the quadratic has no real solutions. Thus, the equation does not have any real roots.