# Solve Equations of the Quadratic Form - Tutorial

 This is a tutorial on solving equations which may be reduced to quadratic form. Detailed solutions and explanations are included. ReviewA quadratic equation has the form ax 2 + bx + c = 0 with a not equal to 0. There are several methods to solve quadratic equations. In this tutorial we use the method of the quadratic formula and the method of factoring. Example 1 : Find all real solutions to the equation. x 4 + x 2 - 6 = 0 Solution to Example 1: Given x 4 + x 2 - 6 = 0 Since (x 2) 2 = x 4, let u = x2 and rewrite the equation in term of u. u2 + u - 6 = 0 Factor the left side. (u + 3)(u - 2) = 0 Use the zero factor theorem to obtain simple equations. a) (u + 3) = 0 b) u - 2 = 0 Solve equation a). u = -3 Solve equation b). u = 2 Use the fact that u = x 2, the first solution in u gives, x 2 = -3 and the second solution gives. x 2 = 2 The square of a real number cannot be negative and therefore the equation x 2 = -3 does not have any real solutions. The second equation is solved by extracting the square root and gives two solutions. x = sqrt(2) x = -sqrt(2) Check Solutions x = sqrt(2) Left side of the equation = sqrt(2) 4 + sqrt(2) 2 - 6 = 4 + 2 - 6 = 0 Right side of the equation = 0. x = -sqrt(2) Left side of the equation = (-sqrt(2)) 4 + (-sqrt(2)) 2 - 6 = 4 + 2 - 6 = 0 Right side of the equation = 0. Conclusion: The real solutions to the given equation are sqrt(2) and -sqrt(2) Matched Exercise 1: Find all real solutions to the equation. x 4 - 2 x 2 - 3 = 0 Example 2 : Find all real solutions to the equation 2 x + 3*sqrt(x) = 5 Solution to Example 2: Given 2 x + 3*sqrt(x) = 5 Note that sqrt(x) implies x has to be positive or zero. Since [ sqrt(x) ]2 = x , let u = sqrt(x) and rewrite the equation in term of u. 2u2 + 3u = 5 Rewrite the equation with the right side equal to 0. 2u2 + 3u - 5 = 0 Use the quadratic formula. The discriminant D is given by D = b2 - 4ac = (3)2 - 4(2)(-5) = 49 Use the quadratic formula to write the two solutions as follows. u1 = (-b + sqrt(D)) / 2a and u2 = (-b - sqrt(D)) / 2a Substitute b, D and a by their values. u1 = (-3 + sqrt(49)) / 4 and u2 = (-3 - sqrt(49)) / 4 Simplify the above expressions. u1 = 1     and     u2 = -5/2 We now use the fact that u = sqrt(x) and solve for x. The first solution u1 gives sqrt(x) = 1 Square both sides to obtain x = 1 The second solution u2 gives sqrt(x) = -5/2 This last equation have no real solutions since the square root of a real positive number is must be a real positive number. Check Solutions x = 1 Left Side = 2 (1) + 3*sqrt(1) = 5 Right Side = 5 Conclusion: The real solution to the given equation is x = 1. Matched Exercise 2. Find all real solutions to the equation. x - 3*sqrt(x) - 4 = 0 More references and links on how to Solve Equations, Systems of Equations and Inequalities.