Any quadratic equation of the form, ax² + bx + c = 0 can be solved using the formula x = -b ± √b² - 4ac / 2a. Here a, b, and c are the coefficients of the x², x, and the numeric term respectively.
We have to solve all of the five equations to be able to match the equations with their solutions.
2x² - 8x + 5, here a = 2, b = -8, c = 5. x = -b ± √b² - 4ac / 2a = -(-8) ± √(-8)² - 4(2)(5) / 2(2) = 8 ± √64 - 40/4. 24 can also be written as 4 × 6 and √4 = 2. So x = 8 ± 2√6 / 2×2= 4±√6/2.
2x² - 10x + 3, here a = 2, b = -10, c = 3. x =-b ± √b² - 4ac / 2a =-(-10) ± √(-10)² - 4(2)(3) / 2(4) = 10 ± √100 + 24/4. 124 can also be written as 4 × 31 and √4 = 2. So x = 10 ± 2√31 / 2×2 = 5 ± √31 /2.
2x² - 8x - 3, here a = 2, b = -8, c = -3. x = -b ± √b² - 4ac / 2a = -(-8) ± √(-8)² - 4(2)(-3) / 2(2) = 8 ± √64 + 24/4. 88 can also be written as 4 × 22 and √4 = 2. Sox = 8 ± 2√22 / 2×2 = 4± √22/2.
2x² - 9x - 1, here a = 2, b = -9, c = -1. x = -b ± √b² - 4ac / 2a = -(-9) ± √(-9)² - 4(2)(-1) / 2(2) = 9 ± √81 + 8/4. x = 9 ± √89 / 4.
2x² - 9x + 6, here a = 2, b = -9, c = 6. x = -b ± √b² - 4ac / 2a = -(-9) ± √(-9)² - 4(2)(6) / 2(2) = 9 ± √81 - 48/4. x = 9 ± √33 / 4 .
C=300+1200x-100x. Find x if the total cost is 3,000 3,000 = 300 + 1200x -100x 3000 = 300 + 1100x 2700 = 1100x 2700÷1100=x x=27/11 x = 2 5/11
Round to 3
