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1. Find the roots of x² + 8x + 11 = 0. Solution: x +8x = x + 8x + (x +? = 5 and X1 = *X2 =

1. Find the roots of x² + 8x + 11 = 0. Solution: x +8x = x + 8x + (x +? = 5 and X1 = *X2 =

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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We have Ax2 + Ay2 + Dx + Ey + F = 0, A # 0 as the general form of the equation of the circl y term and F to represent the constant. Arrange the terms of the given product y- term D for the coefficient of x term, E for Use A to represent the coefficients of and The General We expand the standard form (x -h)² + (y- kY = r2, to get x2-2hx +h2 + y2 - 2ky + k? = r2 x2+ y2 - 2hx- 2ky + (h2+ k2-r?) = 0 OUSups bsnste ACTIVITY 2: FIND MY PARTS This activity will enable you to review completing the square. Directions: Choose inside the box the missing part to each solution. 4 1. Find the roots of x +8x + 11 0. Solution: -4- V5 16 x² +8x = x2 + 8x + -11+16 %3D -5 )² = 5 (x + and -4- V27 X1 = X2 = -11 -4 + V5 -4 + V27 9.
Transcribed Image Text:We have Ax2 + Ay2 + Dx + Ey + F = 0, A # 0 as the general form of the equation of the circl y term and F to represent the constant. Arrange the terms of the given product y- term D for the coefficient of x term, E for Use A to represent the coefficients of and The General We expand the standard form (x -h)² + (y- kY = r2, to get x2-2hx +h2 + y2 - 2ky + k? = r2 x2+ y2 - 2hx- 2ky + (h2+ k2-r?) = 0 OUSups bsnste ACTIVITY 2: FIND MY PARTS This activity will enable you to review completing the square. Directions: Choose inside the box the missing part to each solution. 4 1. Find the roots of x +8x + 11 0. Solution: -4- V5 16 x² +8x = x2 + 8x + -11+16 %3D -5 )² = 5 (x + and -4- V27 X1 = X2 = -11 -4 + V5 -4 + V27 9.
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