Form a quadratic equation with integer coefficients given its roots are -2.5 and 3.

Step 1: Determine the sum of the roots. The given roots are x_1 = -2.5 and x_2 = 3. Sum of roots = x_1 + x_2 = -2.5 + 3 = 0.5 Step 2: Determine the product of the roots. Product of roots = x_1 × x_2 = (-2.5) × 3 = -7.5 Step 3: Form the quadratic equation. A quadratic equation can be written in the form x^2 - (sum of roots)x + (product of roots) = 0. Substituting the sum and product: x^2 - (0.5)x + (-7.5) = 0 x^2 - 0.5x - 7.5 = 0 To express the equation with integer coefficients, multiply the entire equation by 2: 2(x^2 - 0.5x - 7.5) = 2(0) 2x^2 - x - 15 = 0 The equation is in the form ax^2 + bx + c = 0. The quadratic equation is 2x^2 - x - 15 = 0. Step 1: Find the principal value for 2x + 10^. The equation is (2x + 10^) = sqrt(3)2. Let = 2x + 10^. The principal value for is (sqrt(3)2) = 60^. Step 2: Determine all possible values for within the relevant range. Since is positive, lies in the first or second quadrant. The general solutions for = sqrt(3)2 are: = 60^ + n · 360^ = (180^ - 60^) + n · 360^ = 120^ + n · 360^ where n is an integer. The given range for x is 0^ ≤ x ≤ 360^. We need to find the range for = 2x + 10^: 2(0^) + 10^ ≤ 2x + 10^ ≤ 2(360^) + 10^ 10^ ≤ ≤ 730^. Now, list the values of within this range: For n=0: = 60^ = 120^ For n=1: = 60^ + 360^ = 420^ = 120^ + 360^ = 480^ For n=2: = 60^ + 2 · 360^ = 780^ (This is outside the range 10^ ≤ ≤ 730^) = 120^ + 2 · 360^ = 840^ (This is outside the range 10^ ≤ ≤ 730^) So, the valid values for 2x + 10^ are 60^, 120^, 420^, 480^. Step 3: Solve for x. Case 1: 2x + 10^ = 60^ 2x = 50^ x = 25^ Case 2: 2x + 10^ = 120^ 2x = 110^ x = 55^ Case 3: 2x + 10^ = 420^ 2x = 410^ x = 205^ Case 4: 2x + 10^ = 480^ 2x = 470^ x = 235^ All these values are within the range 0^ ≤ x ≤ 360^. The solutions are x = 25^, 55^, 205^, 235^. Send me the next one 📸