Solving equations with polynomials – part 2. n² -7n -30 = 0 ( )( )n n 1 · 30 2 · 15 3 · 10 5 · 6 310 + - n + 3 = 0 n – 10 = 0 - 3 + 10 n = -3n = 10 =

Solving equations with polynomials – part 2

n² -7n -30 = 0 ( )( )n n 1 · 30 2 · 15 3 · 10 5 · 6 310 + - n + 3 = 0 n – 10 = 0 - 3 + 10 n = -3n = 10 = 0

10y² + 4 = 14y 10y²- 14y + 4 = 0 5y²- 7y + 2 = 0 2 ( )( )y 5y 1 · 2 12 - - y – 1 = 0 5y – 2 = 0 + 1 + 2 y = 15y = 2 55 y = ²/₅ 1 · 5

Find the zeros of the following function f(x) = x² - 14x + 45 ( )( )x x 1 · 45 3 · 15 5 · 9 59 - - x – 5 = 0 X – 9 = 0 + 5 + 9 x = 5x = 9 = 0

18m² - 32 = 0 9m²- 16 = 0 2 1 · 16 2 · 8 4 · 4 1 · 9 3 · 3 ( )( )3m 44 - + 3m – 4 = 0 3m + 4 = 0 + 4 - 4 3m = 43m = -4 33 m = 1 ⅓ 33 m = -1 ⅓ = 0

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Presentation on theme: "Solving equations with polynomials – part 2. n² -7n -30 = 0 ( )( )n n 1 · 30 2 · 15 3 · 10 5 · 6 310 + - n + 3 = 0 n – 10 = 0 - 3 + 10 n = -3n = 10 ="— Presentation transcript: