![]() ![]() For example, equations such as 2 x 2 + 3 x − 1 = 0 2 x 2 + 3 x − 1 = 0 and x 2 − 4 = 0 x 2 − 4 = 0 are quadratic equations. Solving Quadratic Equations by FactoringĪn equation containing a second-degree polynomial is called a quadratic equation. If there is a limited amount of space and we desire the largest monitor possible, how do we decide which one to choose? In this section, we will learn how to solve problems such as this using four different methods. ![]() Proportionally, the monitors appear very similar. The computer monitor on the left in Figure 1 is a 23.6-inch model and the one on the right is a 27-inch model. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License. We recommend using aĪuthors: Lynn Marecek, MaryAnne Anthony-Smith Use the information below to generate a citation. Then you must include on every digital page view the following attribution: If you are redistributing all or part of this book in a digital format, ![]() Then you must include on every physical page the following attribution: If you are redistributing all or part of this book in a print format, Want to cite, share, or modify this book? This book uses the This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission. We can use the formula s = A s = A to find the length of a side of a square for a given area. A = s A = s 2 Take the square root of both sides. What if we want to find the length of a side for a given area? Then we need to solve the equation for s.Ī = s 2 Take the square root of both sides. The formula A = s 2 A = s 2 gives us the area of a square if we know the length of a side. If we let s be the length of a side of a square, the area of the square is s 2 s 2. A square is a rectangle in which the length and width are equal. W to find the area of a rectangle with length L and width W.Answer the question with a complete sentence. Check the answer in the problem and make sure it makes sense. Solve the equation using good algebra techniques. Translate into an equation by writing the appropriate formula or model for the situation. Name what we are looking for by choosing a variable to represent it. When appropriate, draw a figure and label it with the given information. Read the problem and make sure all the words and ideas are understood. (Both solutions should work.) The solutions are q = 6 and q = 2. q − 6 = 0 q − 2 = 0 q = 6 q = 2 The checks are left to you. 0 = 9 ( q 2 − 8 q + 12 ) 0 = 9 ( q − 6 ) ( q − 2 ) Use the zero product property. 0 = 9 q 2 − 72 q + 108 Factor the right side. 36 q − 72 = 9 q 2 − 36 q + 36 It is a quadratic equation, so get zero on one side. ( 6 q − 2 ) 2 = ( 3 q − 6 ) 2 Simplify, then solve the new equation. q − 2 + 6 q − 2 + 9 = 4 q + 1 There is still a radical in the equation. q − 2 + 3 = 4 q + 1 The radical on the right side is isolated. Q − 2 + 3 = 4 q + 1 The radical on the right side is isolated.
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