Find the equation whose root are 5/6 and -3/4

To find the equation whose roots are (5)/(6) and -(3)/(4), we can use the relationship between the roots and coefficients of a quadratic equation. For a quadratic equation ax^2 + bx + c = 0, or x^2 - (sum of roots)x + (product of roots) = 0 (when a=1). Step 1: Calculate the sum of the roots. Let the roots be r_1 = (5)/(6) and r_2 = -(3)/(4). Sum of roots = r_1 + r_2 = (5)/(6) + (-(3)/(4)) To add these fractions, find a common denominator, which is 12. Sum of roots = (5 × 2)/(6 × 2) - (3 × 3)/(4 × 3) = (10)/(12) - (9)/(12) = (1)/(12) Step 2: Calculate the product of the roots. Product of roots = r_1 × r_2 = (5)/(6) × (-(3)/(4)) Product of roots = -(5 × 3)/(6 × 4) = -(15)/(24) Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 3. Product of roots = -(15 ÷ 3)/(24 ÷ 3) = -(5)/(8) Step 3: Form the quadratic equation. The general form of a quadratic equation with roots r_1 and r_2 is x^2 - (sum of roots)x + (product of roots) = 0. Substitute the sum and product calculated in the previous steps: x^2 - ((1)/(12))x + (-(5)/(8)) = 0 x^2 - (1)/(12)x - (5)/(8) = 0 Step 4: Clear the denominators to obtain an equation with integer coefficients. The least common multiple (LCM) of the denominators 12 and 8 is 24. Multiply the entire equation by 24: 24 (x^2 - (1)/(12)x - (5)/(8)) = 24 × 0 24x^2 - 24 × (1)/(12)x - 24 × (5)/(8) = 0 24x^2 - 2x - (3 × 5) = 0 24x^2 - 2x - 15 = 0 The equation whose roots are (5)/(6) and -(3)/(4) is: 24x^2 - 2x - 15 = 0 Last free one today — make it count tomorrow, or type /upgrade for unlimited.