btract 7 from both sides of the equation. 3X+7-7=31-7, - 3X=24. Again, we want X alone. To do this we can divide by 3 (each side). 3X/3=24/3, = X =8. Still another example: z2 + 1 = 65 (solve for z). Subtract 1 from each side to get z2 alone. z +1 - 1 = 65 - - - 1, z2=64. Take the square root of each side. Thus, More Complex Equations. Solve for R: 0.25(0.54R + 2.45) = 0.24(2.3R - 1.75), 0.135R + 0.7125 = 0.552R - 0.42, -0.552R + 0.135R = -0.42 0.7125, -0.417R = - 1.0325, R = 2.476. Another example: Solve for X. (x+4)(x-3)=(x-9)(x - 2) = x 2 x - 12 - = x2 - 11x+ 18, 12x = 30, x=2.5. B.3 INTRODUCTORY ALGEBRA B.3.4 Quadratic Equations Quadratic equations will have one term with a square (e.g., X2) and they take the form Ax2 + C = 0 (A#0), where A, B, and C are constants (e.g., numbers). A quadratic equation should always be set to 0 before a solution is attempted. For instance, if we have an equation that is 2x2 + 3x = -21, convert this equation to 2x2 + 3x + 21 = 0. We will discuss two methods of solving a quadratic equation: by factoring and by the quadratic formula. Factoring to Solve a Quadratic Equation. Suppose we have the simple relation x2 1 = 0. We remember from above that this factors into (x - 1) (x + 1) = 0. This being the case, at least one of the factors must equal 0. If this is not understood, realize that there is no other way for the equation statement to be true. Keep in mind that anything multiplied by 0 will be 0. So there are two solutions to the equation: X - 1 = 0, thus X = 1 or x + 1 = 0 and X = Proof that these are correct answers is by substituting them in the equation. Solve for X in this example: x2 100x + 2400 = 0. This factors into (x - 40)(x - 60) = 0. We now have two factors: X - 40 and x - 60, whose product is 0. This means that we must have either X - 40 = 0, where X = 40 or X - 60 = 0, and in this case X = 60. We can check our results by substitution that either of these values satisfies the equation. Another example: Solve for X. (x - 3)(x - 2) = 12. Multiply the factors: x2 - 5x