[b^2 - 4ac]^(1/2) = [16 - 8]^(1/2) = 8^(1/2). Therefore: The roots are real, unequal, and irrational. 2. x^2 - 4x + 6 = 0. [b^2 - 4ac]^(1/2) = [16 - 24]^(1/2) = -8^(1/2). Therefore: The roots are imaginary and unequal. 3. x^2 - 4x + 4 = 0. [b^2 - 4ac]^(1/2) = [16 - 16]^(1/2) = 0^(1/2). Therefore: The roots are real, equal, and rational. ~III. To form the quadratic equation when the roots are given.~ Suppose the roots are 3, -7. Then, x = 3, Or, x - 3 = 0, x = -7. x + 7 = 0. ------------------- Multiplying to get a quadratic, (x - 3)(x + 7) = 0. Or, x^2 + 4x - 21 = 0. _Or_, use the sum and product idea developed on the preceding page. The coefficient of x^2 must be unity. Add the roots and change the sign to get the coefficient of x. Multiply the roots to get the independent term. Therefore: The equation is x^2 + 4x - 21 = 0. In the same way, if the roots are [2 + 3^(1/2)]/7, [2 - 3^(1/2)]/7, the equation is x^2 - (4/7)x + 1/49 = 0. Find the sum, the product, and the nature or character of the roots of the following: 1. x^2 - 7x + 12 = 0. 2. 9x^2 - 6x + 1 = 0. 3. x^2 + 2x + 9734 = 0. 4. 16 + 5/x = 17/(x^2). 5. (x - 8)/(x - 3) = x. 6. (x + 7)(x - 6) = 70. 7. x^2 - x(2)^(1/2) = 3. 8. pr^2 + qr + s = 0. Form the equations whose roots are: 9. 5, -3. 10. 2/3, 5/3. 11. c + d, c - d. 12. -3, -5. 13. [2 +- -3^(1/2)]/5. 14. 8/3 + (2/3)37^(1/2), 8/3 - (2/3)37^(1/2). 15. [-2 +- -2^(1/2)]/2. 16. Solve x^2 - 3x + 4 = 0. Check by substituting the values of x; then check by finding the sum and the product of the roots. Compare the amount of labor required in each case. 17. Solve (x - 3)(x + 2)(x^2 + 3x - 4) = 0. 18. Is e^(4z) + 2e^(3z) + e^(2z) + 2e^z + 2 + e^(-2z) a perfect square? 19. Find the square root (short method): (x^2 - 1)(x^2 - 3x + 2)(x^2 - x - 2). 20. Solve (1.2x - 1.5)/(1.5) + (.4x + 1)/(.2x - .2) = (.4x + 1)/(.5). 21. The glass of a mirror is 18 inches by 12 inches, and it has a frame of uniform width whose area is equal to that of the glass. Find the width of the frame. OUTLINE OF SIMULTANEOUS QUADRATICS ~Simultaneous Quadratics~ CASE I. One equation linear. The other quadratic. 2x + y = 7, x^2 + 2y^2 = 22. METHOD: Solve for x as in terms of y, or _vice versa_, in the linear and substitute in the quadratic. CASE II. Both equations homogeneous and of the second degree. x^2 - xy + y^2 = 39, 2x^2 - 3xy + 2y^2 = 43. METHOD: Let y = vx, and substitute in both equations. ALTERNATE METHOD: Solve for x in terms of y in one equation and substitute in the other. CASE III.