Find the Quadratic Equation of (1) \( -\frac{3}{4} \& \frac{1}{3} \) (2) \( 4-\sqrt{3} \& \quad 4+\sqrt{3} \)

To find the quadratic equations given the roots, we can use the fact that if \( r_1 \) and \( r_2 \) are the roots of a quadratic equation, then the equation can be expressed in the form: \[ x^2 - (r_1 + r_2)x + (r_1 \cdot r_2) = 0 \] ### (1) Roots: \( -\frac{3}{4} \) and \( \frac{1}{3} \) 1. **Sum of the roots**: \[ r_1 + r_2 = -\frac{3}{4} + \frac{1}{3} \] To add these fractions, we need a common denominator. The least common multiple of 4 and 3 is 12. \[ -\frac{3}{4} = -\frac{9}{12}, \quad \frac{1}{3} = \frac{4}{12} \] Thus, \[ r_1 + r_2 = -\frac{9}{12} + \frac{4}{12} = -\frac{5}{12} \] 2. **Product of the roots**: \[ r_1 \cdot r_2 = -\frac{3}{4} \cdot \frac{1}{3} = -\frac{3}{12} = -\frac{1}{4} \] 3. **Quadratic equation**: Substituting the sum and product into the quadratic formula: \[ x^2 - \left(-\frac{5}{12}\right)x + \left(-\frac{1}{4}\right) = 0 \] This simplifies to: \[ x^2 + \frac{5}{12}x - \frac{1}{4} = 0 \] To eliminate the fractions, multiply through by 12 (the least common multiple of the denominators): \[ 12x^2 + 5x - 3 = 0 \] ### (2) Roots: \( 4 - \sqrt{3} \) and \( 4 + \sqrt{3} \) 1. **Sum of the roots**: \[ r_1 + r_2 = (4 - \sqrt{3}) + (4 + \sqrt{3}) = 8 \] 2. **Product of the roots**: \[ r_1 \cdot r_2 = (4 - \sqrt{3})(4 + \sqrt{3}) = 4^2 - (\sqrt{3})^2 = 16 - 3 = 13 \] 3. **Quadratic equation**: Substituting the sum and product into the quadratic formula: \[ x^2 - (8)x + (13) = 0 \] This simplifies to: \[ x^2 - 8x + 13 = 0 \] ### Summary of Quadratic Equations: 1. For roots \( -\frac{3}{4} \) and \( \frac{1}{3} \): \[ 12x^2 + 5x - 3 = 0 \] 2. For roots \( 4 - \sqrt{3} \) and \( 4 + \sqrt{3} \): \[ x^2 - 8x + 13 = 0 \]