Objectives. Solve equations by writing them in quadratic form.. Solve equations that are quadratic in form by using substitution. Vocabulary Prior Knowledge Solve rational equations: Clear the fraction. Watch for restrictions. Solve radical equations. Use the power rule. Check all solutions. New Concepts. Solve equations by rewriting them in quadratic form. Many of the rational and radical equations that we worked with previously result in quadratic equations that can be solved using one of the four methods you have learned. Eample : y 0 0 y y Factor the denominators to find the LCD. LCD = (y )(y + ) y, - 0 y y y y y y y 0 y y Clear the fractions. y y 0 Distribute and foil. Foil on the right and then multiply by -. y y y 0 0 Finish solution. y 0y y 0 0 Check solutions against the restrictions. The solution is -0. is an etraneous solution. Eample : The denominators are all prime. LCD = ( + )( ), V. Zabrocki 0 page
Clear the fractions. 0 0 7 0 0 7 0 Distribute and foil. Divide through by the GCF. 7 This quadratic equation can be factored but if you don t recognize that you can use the Separate the ± to calculate the two solutions. and or. Check solutions against the restrictions. The solutions are and.. Eample : 0 0 Rewrite without negative eponents. Find the LCD: Restrictions: 0. 0 Clear the fractions. 0 This quadratic equation can be factored. ( + )( + ) = 0 + = 0 + = 0 = - = - Check the solutions against the restrictions. The solutions are - and -. V. Zabrocki 0 page
Eample : 0 Isolate the radical term. Use the power rule to remove the radical. Rewrite and foil on the left. Square both the coefficient and the radical on the right. 0 0 0 This equation factors but if you didn t see that, use the ALWAYS check the solutions to radical equations. True 0 0 0 0 0 0 False The solution is.. Solve equations that are quadratic in form using substitution. Any time a given trinomial equation has a first term whose degree is twice the degree of the second term and the third term is a constant, substitution is a good method of solution. There will always be two substitutions: the first to make the problem simpler and the second to solve for the original variable. V. Zabrocki 0 page
Eample : 0 u u 0 Since ( ) appears in both the first term and the second term, once to the second power and once to the first power, substitution is a good method. Let u = ( ), (u + )(u ) = 0 u = - u = = - = = - = Solve this simpler Substitute a second time to find the solutions to the The solutions are - and. Eample : 0 u u 0 Since the degree of the first term is twice the degree of the second term and the third term is a constant, substitution is a good method. Let u =, the variable factor of the middle term. u u 0 Solve this simpler u u Substitute a second time to find the solutions to the Use the square root principle. The solutions are ± and ±. We epect solutions because this is a th degree / / Eample 7: 0 u u 0 Since the degree of the first term is twice the degree of the second term and the third term is a constant, substitution is a good method. Let u = /, the variable factor of the middle term. u u 0 Solve this simpler Factor or use the u u / / Substitute a second time to find the solutions to the V. Zabrocki 0 page
/ / 7 Since / is another way to write the cube root of, we use the power rule to remove the radical. Raise each side to the rd power. The solutions are -/7 and. V. Zabrocki 0 page