![]() The two real solutions are x = 2 and x = -1. The factored (x^3 - 8) and (x^3 + 1) terms can be recognized as the difference of cubes. Now that the substituted values are factored out, replace the u with the original x^3. Here, it would be a lot easier if the expression for factoring was x^2 - 7x - 8 = 0.įirst, let u = x^3, which leaves the factor of u^2 - 7u - 8 = 0. This same strategy can be followed to solve similar large-powered trinomials and binomials.įactor the binomial x^6 - 7x^3 - 8 = 0. Solving each of these terms yields the solutions x = \pm 3, \pm 2. ![]() This is done using the difference of squares equation: a^2 - b^2 = (a + b)(a - b).įactoring (x^2 - 9)(x^2 - 4) = 0 thus leaves (x - 3)(x + 3)(x - 2)(x + 2) = 0. To complete the factorization and find the solutions for x, then (x^2 - 9)(x^2 - 4) = 0 must be factored once more. Once the equation is factored, replace the substitutions with the original variables, which means that, since u = x^2, then (u - 9)(u - 4) = 0 becomes (x^2 - 9)(x^2 - 4) = 0. Now substitute u for every x^2, the equation is transformed into u^2-13u+36=0. There is a standard strategy to achieve this through substitution.įirst, let u = x^2. Here, it would be a lot easier when factoring x^2 - 13x + 36 = 0. Since the discriminant is 0, there is 1 real solution to the equation.Solve for x in x^4 - 13x^2 + 36 = 0.įirst start by converting this trinomial into a form that is more common. Since the discriminant is negative, there are 2 complex solutions to the equation.Ī = 9, b = −6, c = 1 a = 9, b = −6, c = 1 Since the discriminant is positive, there are 2 real solutions to the equation.Ī = 5, b = 1, c = 4 a = 5, b = 1, c = 4 The equation is in standard form, identify a, b, and c.Ī = 3, b = 7, c = −9 a = 3, b = 7, c = −9 ![]() To determine the number of solutions of each quadratic equation, we will look at its discriminant. The left side is a perfect square, factor it.Īdd − b 2 a − b 2 a to both sides of the equation.ĭetermine the number of solutions to each quadratic equation.
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