1 Can Quadratic Techniques Solve Polynomial Equations? PROBLEM 1 Standards PROBLEM 3 PROBLEM 2 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved END SHOW PROBLEM 5 PROBLEM 4 Rational Exponents: Review PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

2 STANDARD 3: Students are adept at operations on polynomials, including long division. STANDARD 8: Students solve and graph quadratic equations by factoring, completing the square, or using the quadratic formula. Students apply these techniques in solving word problems. They also solve quadratic equations in the complex number system. STANDARD 12: Students know the laws of fractional exponents, understand exponential functions, and use these functions in problems involving exponential growth and decay. ALGEBRA II STANDARDS THIS LESSON AIMS: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

3 Estándar 3: Los estudiantes están capacitados para realizar operaciones en polinomios, incluyendo división larga. Estándar 8: Los estudiantes resuelven y grafican ecuaciones mediante factorización, completando el cuadrado, o usando la fórmula cuadrática. Los estudiantes aplican estas técnicas en solución de problemas escritos. Ellos también resuelven ecuaciones en el sistema de números complejos. Estándar 12: Los estudiantes conocen las leyes de los exponentes fraccionarios, entienden funciones exponenciales, y usan estas funciones en problemas que involucran crecimiento o decrecimiento exponencial. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

Standards RATIONAL EXPONENTS: REVIEW X a b = X a 1 b a b X X X Rewrite in radical form: Rational exponents comply with all the exponents’ rules: X = X X 2 3 X – = X 15 = X – 9 X = X = X 20 9 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

Standards Solve the following equation: x –53x = (x ) –53(x ) = x – 53x = 0 2 (-1)(-196) = –53 (-2)(-98) = -100 (-4)(-49) = -53 (x – 4)(x – 49) = (x+2)(x-2)(x+7)(x-7) = 0 x + 2 = 0 x– 2 = 0 x + 7 = 0 x – 7 = 0 –2 x = –2 +2 x = 2 –7 x = –7 +7 x = 7 -7, -2, 2, 7 Rewriting it as a second degree equation: Solving by factoring: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

Standards Solve the following equation: x + 27 = a + b 3 3 = (a + b)(a – ab + b ) 2 2 Recall: x + (3) = x + 3 = 0 –3 x = –3 x –3x + 9 = 0 2 (x + 3)(x –3x + (3) ) = (x + 3)(x –3x + 9) = 0 2 Using the quadratic equation: a= 1 b= -3 c= 9 x= -( ) ( ) - 4( )( ) 2( ) 2 +_ x= 3 9 – ( 4 )( 9 ) 2 +_ x= 2 +_ 3 ( -1)(9)(3) x= 2 +_ x= 2 +_ 3 ( i )(3) 3 x= 2 +_ i x= 2 3 – 3 3 i x= 2 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

Standards Solve the following equation: x – 8 = a – b 3 3 = (a – b)(a + ab + b ) 2 2 Recall: x – (2) = x – 2 = 0 +2 x = 2 x +2x + 4 = 0 2 (x – 2)(x +2x + (2) ) = (x – 2)(x +2x + 4) = 0 2 Using the quadratic equation: a= 1 b= 2 c= 4 x= -( ) ( ) - 4( )( ) 2( ) 2 +_ x= – 2 4 – ( 4 )( 4 ) 2 +_ – x= 2 +_ – 2 ( -1)(4)(3) x= 2 +_ – x= 2 +_ – 2 ( i )(2) 3 x= 2 +_ – 1 + i 3 x= – 1 – i 3 x= PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

Standards Solve the following equation: X +X – 2 = (-1)(2) = 1 –2 1 X – 1= X = X + 2= –2 X = – X = (-2)(-2)(-2) 3 3 X = –8 (X – 1)(X + 2) = PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

Standards Solve the following equation: X – 4 X + 4 = Recall: a – 2ab + b 2 2 = (a – b) 2 X – 2 = 0 2 | X – 2 | = 0 X – 2 = 0 +2 X = X = 4 X – 2 = 0 2 X – 2(2) X + (2) = PRESENTATION CREATED BY SIMON PEREZ. All rights reserved