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On this page we hope to clear up problems you might be having with quadratic equations.  Quadratic equations, or equations of the second degree, such as x2 + 2x - 5 are probably the most common equation you will see in Algebra II (intermediate algebra).  Scroll down or click any of the links below to get a better understanding of quadratic equations. Solving quadratic equations Quadratic formula Quadratic form Quiz on Quadratic Equations Any equation of type ax2 + bx + c = 0 where a, b, and c are constants and a <> 0, is in standard form for a quadratic equation. Quadratic equations of type ax2 + bx + c = 0 and ax2 + bx = 0 (c is 0) can be factored to solve for x.  Examples: 1. Problem: Solve 3x2 + x - 2 = 0 for x. Solution: Factor. (3x - 2)(x + 1) = 0 Use the principle of zero products, which says, if ab = 0, either a, b, or both must be equal to zero. 3x - 2 = 0, x + 1 = 0 3x = 2 , x = -1 x = (2/3) x = -1, (2/3) 2. Problem: Solve 3x2 + 5x = 0 for x. Solution: Factor. x(3x + 5) = 0 Use the principle of zero products. x = 0, 3x + 5 = 0 3x = -5 x = -(5/3) x = 0, -(5/3) Quadratic equations of type ax2 + c = 0 can be solved by solving for x.  Example: 3. Problem: Solve 3x2 = 6 for x. Solution: Recognize that the equation is quadratic because it is the same as 3x2 - 6 = 0. Divide each side by 3. x2 = 2 Take the square root of each side. x = SQRT(2), -(SQRT(2)) Back to Top  Many times you will come across quadratic equations that are not easy to factor or solve.  In those cases, there is a special formula called the quadratic formula that you can use to solve any quadratic equation. The solutions of any quadratic equation, ax2 + bx + c = 0 is given by the following formula, called the quadratic formula: -b ± SQRT(b2 - 4ac) x = ------------------- 2a Example: 1. Problem: Solve 3x2 + 5x = -1 for x. Solution: First find the standard form of the equation and determine a, b, and c. 3x2 + 5x + 1 = 0 a = 3 b = 5 c = 1 Plug the values you found for a, b, and c into the quadratic formula. -5 ± SQRT(52 - 4(3)(1)) x = ----------------------- 2 * 3 Perform any indicated operations. -5 ± SQRT(25 - 12) x = ------------------ 6 -5 ± SQRT(13) x = ------------- 6 The solutions are as follows: -5 + SQRT(13) -5 - SQRT(13) x = -------------, ------------- 6 6 Back to Top  Some equations are not quadratic equations, but are in the same form, such as x4 - 9x2 + 8 = 0.  To solve equations such as that, you make a substitution, solve for the new variable, and then solve for the original variable.  Example: 1. Problem: Solve x4 - 9x2 + 8 = 0 for x. Solution: Let u = x2. Then substitute u for every x2 in the equation. u2 - 9u + 8 =0 Factor. (u - 8)(u - 1) = 0 Utilize the principle of zero products. u - 8 = 0, u - 1 = 0 u = 8 , u = 1 Now substitute x2 for u and solve the equations. x2 = 8, x2 = 1 x = ±SQRT(8), x = ± 1 x = ±2(SQRT(2)) x = ±2(SQRT(2)), ±1 Back to Top  Take the Quiz on quadratic equations.  (Very useful to review or to see if you've really got this topic down.)  Do it!